Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of s characters. The first participant types one character in v1 milliseconds and has ping t1 milliseconds. The second participant types one character in v2 milliseconds and has ping t2 milliseconds.

If connection ping (delay) is t milliseconds, the competition passes for a participant as follows:

  1. Exactly after t milliseconds after the start of the competition the participant receives the text to be entered.
  2. Right after that he starts to type it.
  3. Exactly t milliseconds after he ends typing all the text, the site receives information about it.

The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.

Given the length of the text and the information about participants, determine the result of the game.

Input

The first line contains five integers s, v1, v2, t1, t2 (1 ≤ s, v1, v2, t1, t2 ≤ 1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.

Output

If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".

Examples
Input
5 1 2 1 2
Output
First
Input
3 3 1 1 1
Output
Second
Input
4 5 3 1 5
Output
Friendship
Note

In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.

In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.

In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw.


  题目大意 两个人开始打字游戏,每个人有一个速度,以及发送打字的文本串和提交文本串会有一个延时。问谁会赢。

  按照题目意思,算出运营商收到两人的提交文本最少要耗的时间。

Code

 /**
* Codeforces
* Problem#835A
* Accepted
* Time:30ms
* Memory:2100k
*/
#include <bits/stdc++.h>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
typedef bool boolean;
const signed int inf = (signed)((1u << ) - );
const signed long long llf = (signed long long)((1ull << ) - );
const double eps = 1e-;
const int binary_limit = ;
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
template<typename T>
inline boolean readInteger(T& u){
char x;
int aFlag = ;
while(!isdigit((x = getchar())) && x != '-' && x != -);
if(x == -) {
ungetc(x, stdin);
return false;
}
if(x == '-'){
x = getchar();
aFlag = -;
}
for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
ungetc(x, stdin);
u *= aFlag;
return true;
} int s, v0, v1, t0, t1; inline void init() {
scanf("%d%d%d%d%d", &s, &v0, &v1, &t0, &t1);
} inline void solve() {
int a = * t0 + s * v0;
int b = * t1 + s * v1;
if(a > b)
puts("Second");
else if(a < b)
puts("First");
else
puts("Friendship");
} int main() {
init();
solve();
return ;
}

Codeforces Round #427 (Div. 2) Problem A Key races (Codeforces 835 A)的更多相关文章

  1. Codeforces Round #427 (Div. 2) Problem D Palindromic characteristics (Codeforces 835D) - 记忆化搜索

    Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th nu ...

  2. Codeforces Round #427 (Div. 2) Problem C Star sky (Codeforces 835C) - 前缀和

    The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinat ...

  3. 【Codeforces Round #427 (Div. 2) A】Key races

    [Link]:http://codeforces.com/contest/835/problem/A [Description] [Solution] 傻逼题. [NumberOf WA] [Revi ...

  4. Codeforces Round #427 (Div. 2) Problem B The number on the board (Codeforces 835B) - 贪心

    Some natural number was written on the board. Its sum of digits was not less than k. But you were di ...

  5. Codeforces Round #425 (Div. 2) Problem C Strange Radiation (Codeforces 832C) - 二分答案 - 数论

    n people are standing on a coordinate axis in points with positive integer coordinates strictly less ...

  6. CodeForces 835C - Star sky | Codeforces Round #427 (Div. 2)

    s <= c是最骚的,数组在那一维开了10,第八组样例直接爆了- - /* CodeForces 835C - Star sky [ 前缀和,容斥 ] | Codeforces Round #4 ...

  7. CodeForces 835D - Palindromic characteristics | Codeforces Round #427 (Div. 2)

    证明在Tutorial的评论版里 /* CodeForces 835D - Palindromic characteristics [ 分析,DP ] | Codeforces Round #427 ...

  8. Codeforces Round #716 (Div. 2), problem: (B) AND 0, Sum Big位运算思维

    & -- 位运算之一,有0则0 原题链接 Problem - 1514B - Codeforces 题目 Example input 2 2 2 100000 20 output 4 2267 ...

  9. Codeforces Round #427 (Div. 2)—A,B,C,D题

    A. Key races 题目链接:http://codeforces.com/contest/835/problem/A 题目意思:两个比赛打字,每个人有两个参数v和t,v秒表示他打每个字需要多久时 ...

随机推荐

  1. 39.css3----button按钮点击时出现蓝色边框

    css控制Button 按钮的点击时候出现蓝色边框http://www.inbeijing.org/archives/1139 Button 按钮的点击时候出现蓝色边框的问题 添加css属性,这样在点 ...

  2. 获取PC硬件硬件序列号,唯一标识一台PC

    用一个库:jydisk.dll  百度说是windows提供的C++标准动态库,很方便使用. 点击连接下载文件,里面有各种语言的调用例子,可直接使用.测试多台机器,结果靠谱.没有遇到获取出来是全零的情 ...

  3. bat运行时自己隐藏黑框,而不是用vbs来调用自己

    //autoStart.bat @echo off if "%1" == "h" goto begin mshta vbscript:createobject( ...

  4. LeetCode104.二叉树最大深度

    给定一个二叉树,找出其最大深度. 二叉树的深度为根节点到最远叶子节点的最长路径上的节点数. 说明: 叶子节点是指没有子节点的节点. 示例:给定二叉树 [3,9,20,null,null,15,7], ...

  5. HTML-CSS线性渐变

    实现背景的渐变可以通过为背景添加颜色渐变的图片,也可以使用浏览器的功能来为背景添加渐变的颜色 在IE6或IE7浏览器下可以使用一下示例的CSS语句,设置filter属性来实现颜色 filter:pro ...

  6. html utf-8 中文乱码

    刚才用ajax从记事本中读文档的时候,发现在页面上显示是乱码. 页面编码:<meta charset="utf-8"> 搞半天最后发现是记事本编码格式的问题,记事本默认 ...

  7. 在linux中配置mysql的过程

    以华为企业云中的CentOS 7 为例. 1. 首先要安装相应的包,这个使用各种linux发行版的包管理工具就行,不再赘述.有一点需要注意,现在许多linux发行版将mariadb作为MySQL的默认 ...

  8. caffe-ssd的GPU安装时make runtest报错: BatchReindexLayerTest/3.TestGradient, where TypeParam = caffe::GPUDevice<double>

    报错原因:装了两个cuda,BatchReindexLayerTest/3.TestGradient不能确定用那个 解决办法1:删除其中一个(最好删除9.1,TensorFlow支持的是9.0,为了后 ...

  9. [6]Windows内核情景分析 --APC

    APC:异步过程调用.这是一种常见的技术.前面进程启动的初始过程就是:主线程在内核构造好运行环境后,从KiThreadStartup开始运行,然后调用PspUserThreadStartup,在该线程 ...

  10. 20165215 2017-2018-2 《Java程序设计》第6周学习总结

    20165215 2017-2018-2 <Java程序设计>第6周学习总结 教材学习内容总结 chapter8 Java把String类定义为final类,即String类不能有子类 用 ...