Pashmak decided to give Parmida a pair of flowers from the garden. There are nflowers in the garden and the i-th of them has a beauty number bi. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!

Your task is to write a program which calculates two things:

  1. The maximum beauty difference of flowers that Pashmak can give to Parmida.
  2. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.

Input

The first line of the input contains n (2 ≤ n ≤ 2·105). In the next line there are n space-separated integers b1b2, ..., bn (1 ≤ bi ≤ 109).

Output

The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.

Example

Input

Output
 
Input

Output
 
Input

Output
 

Note

In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:

  1. choosing the first and the second flowers;
  2. choosing the first and the fifth flowers;
  3. choosing the fourth and the second flowers;
  4. choosing the fourth and the fifth flowers.

my code is folowing.

#include <iostream>
using namespace std;
int main()
{
    long long n;
    while( cin >> n )
    {
        int i;
        long long *b=new long long[n];
        for(i=0;i<n;i++)
            cin>>b[i];

        long long maxn=b[0],minn=b[0],m=0,k=0;
        for(i=0;i<n;i++)
        {
            if(b[i]>maxn)
                maxn=b[i];
        }
        for(i=0;i<n;i++)
        {
            if(b[i]<minn)
                minn=b[i];
        }
        for(i=0;i<n;i++)
        {
            if(b[i]==maxn)
                m=m+1;
            if(b[i]==minn)
                k=k+1;
        }
       cout<<(maxn-minn)<<" "<<(maxn==minn)?(n*(n-1)/2):(m*k);
    }

    return 0;
}

  

Pashmak and Flowers的更多相关文章

  1. CF459B Pashmak and Flowers (水

    Pashmak and Flowers Codeforces Round #261 (Div. 2) B. Pashmak and Flowers time limit per test 1 seco ...

  2. cf459B Pashmak and Flowers

    B. Pashmak and Flowers time limit per test 1 second memory limit per test 256 megabytes input standa ...

  3. Codeforces Round #261 (Div. 2) B. Pashmak and Flowers 水题

    题目链接:http://codeforces.com/problemset/problem/459/B 题意: 给出n支花,每支花都有一个漂亮值.挑选最大和最小漂亮值得两支花,问他们的差值为多少,并且 ...

  4. codeforces 459 B.Pashmak and Flowers 解题报告

    题目链接:http://codeforces.com/problemset/problem/459/B 题目意思:有 n 朵 flowers,每朵flower有相应的 beauty,求出最大的beau ...

  5. New Training Table

          2014_8_15 CodeForces 261 DIV2 A. Pashmak and Garden 简单题   B. Pashmak and Flowers    简单题   C. P ...

  6. Codeforces Round #261 (Div. 2) B

    链接:http://codeforces.com/contest/459/problem/B B. Pashmak and Flowers time limit per test 1 second m ...

  7. CF 459A && 459B && 459C && 459D && 459E

    http://codeforces.com/contest/459 A题 Pashmak and Garden 化简化简水题,都告诉平行坐标轴了,数据还出了对角线,后面两个点坐标给的范围也不错 #in ...

  8. Codeforces Round #261 (Div. 2)[ABCDE]

    Codeforces Round #261 (Div. 2)[ABCDE] ACM 题目地址:Codeforces Round #261 (Div. 2) A - Pashmak and Garden ...

  9. [codeforces] 暑期训练之打卡题(二)

    每个标题都做了题目原网址的超链接 Day11<Given Length and Sum of Digits...> 题意: 给定一个数 m 和 一个长度 s,计算最大和最小在 s 长度下, ...

随机推荐

  1. 详谈Javascript类与继承

    本文将从以下几方面介绍类与继承 类的声明与实例化 如何实现继承 继承的几种方式 类的声明与实例化 类的声明一般有两种方式 //类的声明 var Animal = function () { this. ...

  2. vue.js的学习中的简单案例

    今天学习了近年来挺火的一门JS技术,叫vue.js下面是它的一个简单案例: <html> <head> <title>$Title$</title> / ...

  3. Foundation框架的小总结

    一.Foundation框架—结构体 一.基本知识 Foundation框架中包含了很多开发中常用的数据类型,如结构体,枚举,类等,是其他ios框架的基础. 如果要想使用foundation框架中的数 ...

  4. 读Kafka Consumer源码

    最近一直在关注阿里的一个开源项目:OpenMessaging OpenMessaging, which includes the establishment of industry guideline ...

  5. PE文件格式分析

    PE文件格式分析 PE 的意思是 Portable Executable(可移植的执行体).它是 Win32环境自身所带的执行文件格式.它的一些特性继承自Unix的Coff(common object ...

  6. weex 环境搭建

    最近为了项目需要(实际上是为了年底KPI),领导要求用3天时间,学习并使用weex开发一个页面,说实话,压力山大.在这之前压根儿就没听说过啊,一脸懵逼 无奈之余只能Google了,惊喜的发现weex的 ...

  7. jsp加java连接数据库,进行信息输入,并进行初步的拦截判断。

    图形大概这样 按照图片要求设计添加新课程界面.(0.5分) 在后台数据库中建立相应的表结构存储课程信息.(0.5分) 实现新课程添加的功能. 要求判断任课教师为王建民.刘立嘉.刘丹.王辉.杨子光五位教 ...

  8. Django学习(2)数据宝库

    数据库是一所大宝库,藏着各种宝贝.一个没有数据库的网站,功能有限.在Django中,支持的数据库有以下四种: SQLite3 MySQL PostgreSQL Oracle 其中SQLite3为Dja ...

  9. NYOJ 138 找球号(二) bitset 二进制的妙用

    找球号(二) 时间限制:1000 ms  |  内存限制:65535 KB 难度:5 描述 描述 在某一国度里流行着一种游戏.游戏规则为:现有一堆球中,每个球上都有一个整数编号i(0<=i< ...

  10. yii2.0中Rbac 怎么添加超加管理员

    最笨的是定义常量.具体怎么做?看下面: //定义在控制器声明上面define('BEST_PHPER',serialize(array('admin','admin1')));//设置admin管理员 ...