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>

#define swap(x,y) do{long t=x; x=y; y=t;}while(0)

void sort(long a[],int left,int right){
	int pl=left;
	int pr=right;
	long x=a[(pl+pr)/2];

	do{
		while(a[pl]<x) pl++;
		while(a[pr]>x) pr--;
		if(pl<=pr){
			swap(a[pl],a[pr]);
			pl++;
			pr--;
		}
	}while(pl<=pr);
	if(left<pr) sort(a,left,pr);
	if(pl<right) sort(a,pl,right);
}

int main(void){
  int i,j;

  int n;
  long k;
  scanf("%d%ld",&n,&k);

  long long all_sum=0;
  long a[n];
  for(i=0;i<n;i++){
    scanf("%ld",&a[i]);
    all_sum+=a[i];
  }

  sort(a,0,n-1);

  int rest;
  long long sum=0;
  for(i=0;;i++){
    if(sum+a[i]<k){
      sum+=a[i];
    }else{
      rest=n-i;
      break;
    }
  }

  for(i=rest;;i++){
    if(all_sum%i==0){
      printf("%lld\n",all_sum/i);
      break;
    }
  }

  return 0;
}