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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>

int main(){
  int n,k,p,div,sum,f;
  int flag=0;
  scanf("%d %d",&n,&k);
    
  int a[n+1];
  for(int i=1;i<=n;i++){
    scanf("%d",&a[i]);
  }
    
  p=0;
  for(int i=1;i<=n;i++){
    p=p+a[i];
  }
    
  for(int i=p;i>1;i--){
    f=0;
    sum=0;
    int c[i];
    if(p%i==0){
      for(int l=0;l<=i-1;l++){
        c[l]=0;
      }
    }
    for(int j=1;j<=n;j++){
      if(p%i!=0){
        break;
      }
      c[a[j]%i]=c[a[j]%i]+1;
      sum=sum+a[j]%i;
    }
    if(p%i==0){
      div=sum/i;
      
      int sum1=0;
      int d,e;
      for(int l=i-1;;l--){
        if(sum1+c[l]<div){
          sum1=sum1+c[l];
        }else{
          d=sum1+c[l]-div;
          e=l;
          break;
        }
      }
      for(int l=0;l<=e-1;l++){
        f=f+c[l]*l;
      }
      f=f+e*d;
      
      if(f<=k){
        printf("%d",i);
        flag=1;
        break;
      }
    }
  }
  
  if(flag==0){
    printf("1");
  }
    
  return 0;
}