Problem:
Score : 500 points

Problem Statement

We have a sequence of N integers: A_1, A_2, \cdots, A_N.

You can perform the following operation between 0 and K times (inclusive):

Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.

Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.

Constraints

2 \leq N \leq 500

1 \leq A_i \leq 10^6

0 \leq K \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_{N-1} A_{N}

Output

Print the maximum possible positive integer that divides every element of A after the operations.

Sample Input 1

2 3
8 20

Sample Output 1

7

7 will divide every element of A if, for example, we perform the following operation:

Choose i = 2, j = 1. A becomes (7, 21).

We cannot reach the situation where 8 or greater integer divides every element of A.

Sample Input 2

2 10
3 5

Sample Output 2

8

Consider performing the following five operations:

Choose i = 2, j = 1. A becomes (2, 6).

Choose i = 2, j = 1. A becomes (1, 7).

Choose i = 2, j = 1. A becomes (0, 8).

Choose i = 2, j = 1. A becomes (-1, 9).

Choose i = 1, j = 2. A becomes (0, 8).

Then, 0 = 8 \times 0 and 8 = 8 \times 1, so 8 divides every element of A. We cannot reach the situation where 9 or greater integer divides every element of A.

Sample Input 3

4 5
10 1 2 22

Sample Output 3

7

Sample Input 4

8 7
1 7 5 6 8 2 6 5

Sample Output 4

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const int maxn=3e5+10;
int n,a[maxn],k;
int sum,ans;
int cur;
vector<int> d;

int main(){
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        sum+=a[i];
    }
    for(int i=(int)sqrt(sum);i>=2;i--){
        if(sum%i==0){
            d.push_back(i);
            if(i*i!=sum){
                d.push_back(sum/i);
            }
        }
    }
    d.push_back(sum);
    for(int x:d){
        vector<int> b;
        for(int i=1;i<=n;i++){
            b.push_back(a[i]%x);
        }
        sort(b.rbegin(),b.rend());
        while(!b.empty()&&b.back()==0){
            b.pop_back();
        }
        reverse(b.begin(),b.end());
        int pt=0;
        cur=0;
        for(int i=0;i<(int)b.size();i++){
            while(b[i]>0){
                if(b.empty())break;
                if(x-b.back()==b[i]){
                    cur+=(x-b.back());
                    b[i]=0;
                    b.pop_back();
                    break;
                }
                if(b[i]>x-b.back()){
                    cur+=(x-b.back());
                    b[i]-=(x-b.back());
                    b.pop_back();
                }
                else if(b[i]<x-b.back()){
                    cur+=(b[i]);
                    b.back()+=b[i];
                    b[i]=0;
                }
            }
        }
        if(cur<=k){
            ans=max(ans,x);
        }
    }
    printf("%d\n",ans);
}
/*
    Good Luck
        -Lucina
*/