D. Robin Hood time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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 c_i 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 ≤ 500 000, 0 ≤ k ≤ 10⁹) — 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 c_i (1 ≤ c_i ≤ 10⁹) — 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

Note

Lets look at how wealth changes through day in the first sample.

1. [1, 1, 4, 2]
2. [2, 1, 3, 2] or [1, 2, 3, 2]

So the answer is 3 - 1 = 2

In second sample wealth will remain the same for each person.

題意：給你n個數，k次操作， 每次操作把最大的減一，最小的加一， 如果最大的減一變成最小的，就吧這減掉的1再加回去

直接暴力900多ms躺過，，，，

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
const int inf = 2147483647;
const double PI = acos(-1.0);
const double e = 2.718281828459;
const int mod = 1000000007;
typedef long long LL;
#pragma comment(linker, "/STACK:102400000,102400000")
//freopen("in.txt","r",stdin); //輸入重定向,輸入資料將從in.txt檔案中讀取
//freopen("out.txt","w",stdout); //輸出重定向,輸出資料將儲存在out.txt檔案中cin
int a[500005];
map<LL, LL>mp;
set<int>st;
set<int>::iterator it, tt;
set<int>::reverse_iterator rt, rtt;

int main()
{
	int n, k, i, j;
	while (cin >> n >> k)
	{
		LL s = 0;
		mp.clear();
		st.clear();
		for (i = 1; i <= n; ++i)
		{
			scanf("%d", &a[i]);
			st.insert(a[i]);
			mp[a[i]]++;
			s += a[i] * 1LL;
		}
		sort(a + 1, a + 1 + n);
		LL x = s / n;
		LL y = s % n;
		LL num = 0;
		for (i = 1; i <= n; ++i)
		{
			LL xx = x;
			if (i >= n - y + 1)
				xx++;
			if (a[i] <= xx)
			{
				num += xx - a[i];
			}
		}
		if (k >= num)//如果k比num大,最後肯定是一種平衡狀態，怎麼操作都不會變了，要麼是0，要麼是1
		{
			if (y == 0)
				printf("0\n");
			else
				printf("1\n");
			
		}
		else
		{
			int k1 = k, k2 = k;
			int mmin, mmax;
			for (it = ++st.begin(); it != st.end(); ++it)//正著跑一邊，看k次操作後的最小值是多少
			{
				tt = --it;
				++it;
				if (mp[*tt] * (*it - *tt) * 1LL > k1 )//注意兩個相乘可能爆int，所以map直接longlong即可
				{
					mmin = (*tt + k1 / mp[*tt]);
					break;
				}
				else if (mp[*tt] * (*it - *tt) * 1LL == k1)
				{
					mmin = *it;
					break;
				}
				else
				{
					k1 -= mp[*tt] * (*it - *tt);
					mp[*it] += mp[*tt];
				}
			}
			for (rt = st.rbegin(); rt != --st.rend(); ++ rt)//反著跑一邊求出最大值
			{
				rtt = ++rt;
				--rt;
				if (mp[*rt] * (*rt - *rtt) * 1LL > k2)
				{
					mmax = *rt - (k2 / mp[*rt]);
					break;
				}
				else if (mp[*rt] * (*rt - *rtt) * 1LL == k2)
				{
					mmax = *rtt;
					break;
				}
				else
				{
					k2 -= mp[*rt] * (*rt - *rtt);
					mp[*rtt] += mp[*rt];
				}
			}
			printf("%d\n", mmax - mmin);
		}


	}
}