题目来源：Robin Hood

Statement

We all know the impressive story of Robin Hood. Robin Hood uses his archery skills and his wits to steal the money from rich, and return it to the poor.

There are n citizens in Kekoland, each person has ci coins. Each day, Robin Hood will take exactly 1 coin from the richest person in the city and he will give it to the poorest person (poorest person right after taking richest's 1 coin). In case the choice is not unique, he will select one among them at random. Sadly, Robin Hood is old and want to retire in k days. He decided to spend these last days with helping poor people.

After taking his money are taken by Robin Hood richest person may become poorest person as well, and it might even happen that Robin Hood will give his money back. For example if all people have same number of coins, then next day they will have same number of coins too.

Your task is to find the difference between richest and poorest persons wealth after k days. Note that the choosing at random among richest and poorest doesn't affect the answer.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 500000, 0 ≤ k ≤ 1e9) — the number of citizens in Kekoland and the number of days left till Robin Hood's retirement.

The second line contains n integers, the i-th of them is ci (1 ≤ ci ≤ 1e9) — initial wealth of the i-th person.

Output

Print a single line containing the difference between richest and poorest peoples wealth.

Examples

input

4 1
1 1 4 2

output

2

input

3 1
2 2 2

output

0

题意

给定n个数字和k次操作，每次操作都会使n个数字中的最大值减1最小值加1，问k次操作后最大值与最小值的差是多少。

思路

先求平均值，所有被减1的数字均大于平均值，所有被加1的数字均小于平均值。二分出最终的最大值和最小值，要使得最大值最小，最小值最大。

代码

#include <bits/stdc++.h>
using namespace std;
long long n, k;
long long a[500005];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    while (cin >> n >> k)
    {
        long long sum = 0;
        for (int i = 1; i <= n; ++i)
        {
            cin >> a[i];
            sum += a[i];
        }
        sort(a + 1, a + 1 + n);

        long long l, r, mid, ans1, ans2;

        l = 0, r = sum / n;
        while (l < r)
        {
            mid = (l + r + 1) / 2;
            long long p = 0;
            for (int i = 1; i <= n; ++i)
                if (mid > a[i])
                    p += mid - a[i];
            if (p <= k)
                l = mid;
            else
                r = mid - 1;
        }
        ans1 = l;

        l = sum / n + !(sum%n == 0), r = a[n];
        while (l < r)
        {
            mid = (l + r) / 2;
            long long p = 0;
            for (int i = 1; i <= n; ++i)
                if (mid < a[i])
                    p += a[i] - mid;
            if (p <= k)
                r = mid;
            else
                l = mid + 1;
        }
        ans2 = l;
        cout << ans2 - ans1 << endl;
    }
    return 0;
}

注意

1. mid = (l + r + 1) / 2; r = mid - 1; 最小值最大化
   mid = (l + r) / 2; l = mid + 1; 最大值最小化
2. 该题求最大值的时候的取值范围
   如果平均值是整数为[average, max]
   如果不是整数则为[average+1, max]