codeforces 672D 二分

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 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 ≤ 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 ci (1 ≤ ci ≤ 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

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： 有n個人， 每個人都有一定的财富值， 每天有最多财富的人會把自己的一進制錢給最少财富的人，求k天之後最富有的人跟最少财富的人的內插補點是多少。

分析： 可以二分的查找最富有的人的錢數和最少财富的人的錢數，然後兩者相減就是結果，在二分之前還有确定最多的錢的下界和最少的錢的上界。如果總錢數sum能整除n，那麼上界跟下界都可以取到sum/n，否則上界能取到sum/n, 下界能取到sum/n加1。有了上下界， 就可以進行二分了。

#include <bitset>
#include <map>
#include <vector>
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#define inf 0x3f3f3f3f
#define mem(a,x) memset(a,x,sizeof(a))
#define F first
#define S second
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;

inline int in()
{
    int res=0;char c;int f=1;
    while((c=getchar())<'0' || c>'9')if(c=='-')f=-1;
    while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
    return res*f;
}
const int N=510000,MOD=1e9+7;
int a[N],n,k;
bool check1(int x)
{
    ll sum=0;
    for(int i=0;i<n;i++){
        if(a[i]<x) sum += x-a[i];
    }
    return sum <= k;
}

bool check2(int x)
{
    ll sum=0;
    for(int i=0;i<n;i++){
        if(a[i]>x) sum += a[i]-x;
    }
    return sum <= k;
}
int main()
{
    n=in(),k=in();
    ll sum=0;
    for(int i=0;i<n;i++){
        a[i]=in();
        sum += a[i];
    }
    ll minup,maxdown;
    if(sum % n == 0){
        minup = maxdown = sum/n;
    }
    else{
        minup = sum/n;
        maxdown = minup+1;
    }
    ll l=1,r=minup;
    ll ans;
    while(l<=r){
        int mid = l+r>>1;
        if(check1(mid)) l=mid+1,ans=mid;
        else r=mid-1;
    }
    l=maxdown,r=1e9;
    ll tmp;
    while(l<=r){
        int mid=l+r>>1;
        if(check2(mid)) r=mid-1,tmp=mid;
        else l=mid+1;
    }
    ans = tmp-ans;
    cout<<ans<<endl;
    return 0;
}