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: Accepted)

#include<bits/stdc++.h>
using namespace std;

int main(){
  int n,k,d,l,r,var,count,sum=0,ans=1;
  cin>> n >> k;
  vector<int> divide,a(n),remain(n);
  for (int i=0; i<n; i++){
    cin>> a[i];
    sum+=a[i];
  }
  
  for (int i=1; i*i<=sum; i++){
    if (sum%i==0){
      divide.push_back(i);
      divide.push_back(sum/i);
    }
  }
  sort(divide.begin(),divide.end()); //総和の約数を昇順に
  
  for (int i=divide.size()-1; i>0; i--){ //i=0はd=1になって必ず答えになる。あらかじめans=1にして省略
    d=divide[i];
    for (int j=0; j<n; j++) remain[j]=a[j]%d;
    sort(remain.begin(),remain.end()); //余りの配列を昇順に
    
    count=0; l=0; r=n-1;
    while (l<r){ //操作k回以内にremain配列を0とdだけにできるか
      if (remain[l]==0){ l++; continue;}
      if (remain[r]==d){ r--; continue;}
      var=min(remain[l],d-remain[r]);
      count+=var;
      if (k<count) break;
      remain[l]-=var; remain[r]+=var;
    }
    if (remain[l]%d==0){
      ans=d;
      break;
    }
  }
  
  cout<< ans << endl;
}