There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.

Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given n, print the last digit of 1378n.

Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.

Input

The single line of input contains one integer n (0  ≤  n  ≤  109).

Output

Print single integer — the last digit of 1378n.

Examples

Input
1
Output
8
Input
2
Output
4

Note

In the first example, last digit of 13781 = 1378 is 8.

In the second example, last digit of 13782 = 1378·1378 = 1898884 is 4.

#include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
if(n==)
printf("1\n");
else if(n%==)
{
printf("8\n");
}
else if(n%==)
{
printf("4\n");
}
else if(n%==)
{
printf("2\n");
}
else
{
printf("6\n");
}
return ;
}

Vladik is a competitive programmer. This year he is going to win the International Olympiad in Informatics. But it is not as easy as it sounds: the question Vladik face now is to find the cheapest way to get to the olympiad.

Vladik knows n airports. All the airports are located on a straight line. Each airport has unique id from 1 to n, Vladik's house is situated next to the airport with id a, and the place of the olympiad is situated next to the airport with id b. It is possible that Vladik's house and the place of the olympiad are located near the same airport.

To get to the olympiad, Vladik can fly between any pair of airports any number of times, but he has to start his route at the airport a and finish it at the airport b.

Each airport belongs to one of two companies. The cost of flight from the airport i to the airport j is zero if both airports belong to the same company, and |i - j| if they belong to different companies.

Print the minimum cost Vladik has to pay to get to the olympiad.

Input

The first line contains three integers n, a, and b (1 ≤ n ≤ 105, 1 ≤ a, b ≤ n) —
the number of airports, the id of the airport from which Vladik starts
his route and the id of the airport which he has to reach.

The second line contains a string with length n, which consists only of characters 0 and 1. If the i-th character in this string is 0, then i-th airport belongs to first company, otherwise it belongs to the second.

Output

Print single integer — the minimum cost Vladik has to pay to get to the olympiad.

Examples

Input
4 1 4
1010
Output
1
Input
5 5 2
10110
Output
0

Note

In the first example Vladik can fly to the airport 2 at first and pay |1 - 2| = 1 (because the airports belong to different companies), and then fly from the airport 2 to the airport 4 for free (because the airports belong to the same company). So the cost of the whole flight is equal to 1. It's impossible to get to the olympiad for free, so the answer is equal to 1.

In the second example Vladik can fly directly from the airport 5 to the airport 2, because they belong to the same company.

 #include<stdio.h>
int main()
{
int n,a,b;
char s[];
scanf("%d %d %d",&n,&a,&b);
scanf("%s",s);
if(s[a-]==s[b-])
printf("0\n");
else
printf("1\n");
return ;
}

Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.

Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (n - 1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1, 2, 1] after the first step, the sequence [1, 2, 1, 3, 1, 2, 1] after the second step.

The task is to find the value of the element with index k (the elements are numbered from 1) in the obtained sequence, i. e. after (n - 1) steps.

Please help Chloe to solve the problem!

Input

The only line contains two integers n and k (1 ≤ n ≤ 50, 1 ≤ k ≤ 2n - 1).

Output

Print single integer — the integer at the k-th position in the obtained sequence.

Examples

Input
3 2
Output
2
Input
4 8
Output
4

Note

In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.

In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.

 #include<bits/stdc++.h>
using namespace std;
long long n,k;
int main()
{
scanf("%lld%lld",&n,&k);
cout<<log2(k&(-k))+<<endl;
}

Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction as a sum of three distinct positive fractions in form .

Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that . Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.

If there is no such answer, print -1.

Input

The single line contains single integer n (1 ≤ n ≤ 104).

Output

If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.

If there are multiple answers, print any of them.

Examples

Input
3
Output
2 7 42
Input
7
Output
7 8 56
 #include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
if(n==)printf("-1\n");
else printf("%d %d %d\n",n,n+,n*(n+));
return ;
}

ACM 第三天的更多相关文章

  1. 【ACM】三点顺序

    三点顺序 时间限制:1000 ms  |  内存限制:65535 KB 难度:3   描述 现在给你不共线的三个点A,B,C的坐标,它们一定能组成一个三角形,现在让你判断A,B,C是顺时针给出的还是逆 ...

  2. 寒假的ACM训练三(PC110107/UVa10196)

    #include <iostream> #include <string.h> using namespace std; char qp[10][10]; int result ...

  3. ACM第三次比赛 Big Chocolate

    Problem G Big Chocolate Mohammad has recently visited Switzerland . As he loves his friends very muc ...

  4. ACM第三次比赛UVA11877 The Coco-Cola Store

      Once upon a time, there is a special coco-cola store. If you return three empty bottles to the sho ...

  5. ACM第三题 完美立方

    形如a3= b3 + c3 + d3的等式被称为完美立方等式.例如123= 63 + 83 + 103 .编写一个程序,对任给的正整数N (N≤100),寻找所有的四元组(a, b, c, d),使得 ...

  6. 今天的第一个程序-南阳acm输入三个数排序

    #include<stdio.h>main(){    int a,b,c,t;    scanf("%d%d%d",&a,&b,&c);    ...

  7. (转)女生应该找一个玩ACM的男生

    1.强烈的事业心 将来,他也一定会有自己热爱的事业.而且,男人最性感的时刻之一,就是他专心致志做事的时候.所以,找一个机会在他全神贯注玩ACM的时候,从侧面好好观察他,你就会发现我说的话没错. 2.永 ...

  8. ACM心情总结

    已经快要12点了,然而还有5000字概率论论文没有动.在论文里,我本来是想要总结一下ACM竞赛中出现过的概率论题目,然而当敲打第一段前言的时候,我就迟疑了. 我问自己,ACM竞赛到底有什么现实意义. ...

  9. 【ACMER纷纷表示】女生应该找一个玩ACM的男生

    1.强烈的事业心 将来,他也一定会有自己热爱的事业.而且,男人最性感的时刻之一,就是他专心致志做事的时候.所以,找一个机会在他全神贯注玩ACM的时候,从侧面好好观察他,你就会发现我说的话没错.2.永不 ...

随机推荐

  1. React Native开发之expo中camera的基本使用

    之前做RN项目没调用过本地摄像头,今天下班早,做了一个简单的小demo:主要实现的功能:点击拍照按钮进入拍照界面,点击flip进行前后摄像头转换,点击开始拍照实现拍照功能(没写保存到本地的功能,大家可 ...

  2. 解决 Node.js 错误 Error:listen EADDRINUSE

    第一次尝试 node.js 中的 express 框架,写了第一个 js 文件之后,在 WebStorm 运行,到游览器刷新,成功运行. 又创建一个 js 文件,写的是静态路由的访问,结果出现了 Er ...

  3. HyperLedger Fabric 1.4 基础环境搭建(7)

    学习了前面几章理论知识后,本章开始介绍实践操作,先介绍Fabric基础环境搭建,采用的操作系统为Centos 7 64位,依次介绍Docker安装.Docker-Compose安装.GO语言环境安装. ...

  4. 20154327 Exp5 MSF基础应用

    基础问题回答 用自己的话解释什么是exploit,payload,encode. exploit漏洞利用,一般出现漏洞后,根据一些大佬们给出的POC尝试去进行漏洞利用. payload攻击负载,是我们 ...

  5. VINS(一)简介与代码结构

    VINS-Mono和VINS-Mobile是香港科技大学沈劭劼团队开源的单目视觉惯导SLAM方案.是基于优化和滑动窗口的VIO,使用IMU预积分构建紧耦合框架.并且具备自动初始化,在线外参标定,重定位 ...

  6. Qml-Dialog不能隐藏标题栏和按钮自定义

    在项目中,需要弹出一个对话框来完成用户输入的功能,为了考虑界面的同一,这里需要将原生自带的标题栏隐藏掉,换成自己写的 按照widget的写法,可以使用QDialog,但是qml与之对应的Dialog我 ...

  7. 负数取余/整除,Python和C语言的不同

    总结一句:Python中负数整除,是向负无穷取整,所以导致负数取余不对 在数学公式中,两种语言的表示算法都是一样的,都是: r=a-n*[a/n] 以上,r是余数,a是被除数,n是除数. 唯一不同点, ...

  8. AirtestIDE实践一:梦幻西游手游师门任务自动化

    Airtest Project是网易自研的游戏自动化项目.Airtest IDE是这个项目的一个IDE,就像Eclipse.Pycharm一样,是一个集成开发工具.Airtest框架是一个基于Open ...

  9. selenium--driver.switchTo()

    在自动化测试中,会遇到多窗口.多iframe.多alert的情况.此时,会使用driver.switchTo()来解决. 下面时关于driver.switchTo()的详细介绍: 1.多windows ...

  10. 七 Appium常用方法介绍

    文本转自:http://www.cnblogs.com/sundalian/p/5629609.html 由于appium是扩展了Webdriver协议,所以可以使用webdriver提供的方法,比如 ...