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 hasc_i coins. Each day, Robin Hood will take exactly1 coin from the richest person in the city and he will give it to the poorest person (poorest person right after taking richest's1 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 ink 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 afterk days. Note that the choosing at random among richest and poorest doesn't affect the answer.

The first line of the input contains two integers n andk (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, thei-th of them is c_i (1?≤?c_i?≤?10⁹) — initial wealth of the i-th person.

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

4 11 1 4 2Output
2Input
3 12 2 2Output
0Note
Lets look at how wealth changes through day in the first sample.
[1,?1,?4,?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.
題目大意：
給你N個人的財富值，每天最有錢的人會給最沒錢的人一塊錢，問K天之后，最窮的人和最富有的人之間差多少錢。
思路：
1、virtual contest過程中一直在二分差值，然后發現最小值和最大值的問題不是很好處理，一直在想一個科學的連續二分的方式去枚舉出最小值和最大值。以一種二分套二分的方式去解，以失敗告終。
2、正解是這樣的：①因為時間越長（天數經過的越多），最大值就會越小，同理，最小值就會越大（所以這就是智障的去二分差值的理由？尼瑪本質是最大值最小值的變化好嘛，為毛要想到差值上去.......）那么我們二分一個最小值，接下來二分一個最大值。
②判定二分的過程很簡單，對于枚舉最小值的時候，如果需要的最少天數小于等于k，那么就加大最小值，否則減小最小值即可。那么二分枚舉最大值的時候同理即可。
③問題的坑點在于維護上下界。對于總值：sum，如果sum%n==0，那么最小值的上界就是sum/n，最大值的下界也是sum/n.當sum%n！=0的時候，最小值的上界還是sum/n，但是最大值的下界應該是sum/n+1.這里被坑了......................
3、過程維護最終解，細心一些沒有別的了。
Ac代碼：
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define ll __int64ll a[500500];ll n,m;int Slove(ll mid){    ll sum=0;    for(int i=1;i<=n;i++)    {        if(a[i]<mid)sum+=mid-a[i];    }    if(sum<=m)return 1;    else return 0;}int Slove2(ll mid){    ll sum=0;    for(int i=n;i>=1;i--)    {        if(a[i]>mid)sum+=abs(mid-a[i]);    }    if(sum<=m)return 1;    else return 0;}int main(){    while(~scanf("%I64d%I64d",&n,&m))    {        ll sum=0;        ll up,down;        for(int i=1;i<=n;i++)        {            scanf("%I64d",&a[i]);            sum+=a[i];        }        if(sum%n==0)up=down=sum/n;        else up=sum/n,down=up+1;        sort(a+1,a+1+n);        ll l=0;        ll r=up;        ll minn=-1;        while(r-l>=0)        {            ll mid=(l+r)/2;            if(Slove(mid)==1)            {                minn=mid;                l=mid+1;            }            else r=mid-1;        }        l=down;        r=1000000000;        ll maxn=-1;        while(r-l>=0)        {            ll mid=(l+r)/2;            if(Slove2(mid)==1)            {                maxn=mid;                r=mid-1;            }            else l=mid+1;        }        printf("%I64d/n",maxn-minn);    }}