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;
    int p=0;
    int flag=0;
    scanf("%d %d",&n,&k);
    
    int a[n+1];
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    
    for(int i=1;i<=n;i++){
        p=p+a[i];
    }
    
    for(int i=p;i>1;i--){
        int f=0;
        int 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){
            int div=sum/i;
            
            int sum1=0;
            int d=0;
            int e=0;
            for(int l=i-1;l>=0;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;
}