codeforces 459B Pashmak and Flowers

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B. Pashmak and Flowers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Pashmak decided to give Parmida a pair of flowers from the garden. There are n flowers 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 b1, b2, ..., 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.

Examples

input

2
1 2

output

1 1

input

3
1 4 5

output

4 1

input

5
3 1 2 3 1

output

2 4

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.

求最大与最小之差和最大与最小的组合数！//注意最大值与最小值相等的情况，组合数为n*(n-1)/2，和int的范围（超出了int）。

#include<stdio.h>
#include<algorithm>
using namespace std;
long long s[200005];
int main()
{
    long long n,i,a=0,b=0;
    long long minn=0x3f3f3f3f;
    long long maxn=-0x3f3f3f3f;
    scanf("%I64d",&n);
    for(i=0; i<n; i++)
    {
        scanf("%I64d",&s[i]);
        if(minn>s[i])
        {
            minn=s[i];
        }
        if(maxn<s[i])
        {
            maxn=s[i];
        }
    }
    if(minn==maxn)
    {
        printf("0 %I64d\n",n*(n-1)/2);
        return 0;
    }
    for(i=0; i<n; i++)
    {
        if(s[i]==minn)
            a++;
        if(s[i]==maxn)
            b++;
    }
    printf("%I64d %I64d\n",maxn-minn,a*b);
    return 0;
}