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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>


int compare(int *a, int *b){
  return *a - *b;
}

int main(){
  int n,k,
      a[500],b[500],sum=0,l,t,y,
      i,j;
  
  scanf("%d%d", &n, &k);
  
  for(i=0;i<n;i++){
	scanf("%d", &a[i]);
    sum += a[i];
  }
  
  l = sum;
  while(1){
    if(sum%l == 0){
      for(i=0;i<n;i++){
        b[i] = a[i] % l;
      }
      qsort(b, n, sizeof(int), compare);
      y=0;
      t=0;
      while(i <= j){
        if(0 >= y){
          y += b[i];
          t += b[i];
          i++;
        }else{
          y -= (l - b[j]);
          j--;
        }
      }
      if(t <= k){
        printf("%d", l);
        return 0;
      }
    }
    l--;
  }
}