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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int n, k, sum, a[505], b[505], c[505];
int comp(int a, int b){
	return a>b;
}
int main(){
	cin>>n>>k;
	for(int i=1;i<=n;i++){
		scanf("%d", &a[i]);
		sum+=a[i];
	}
	//sort(a+1, a+n+1);
	for(int i=1;i*i<=sum;i++){
		if(sum%i==0){
			int p=sum/i, now=0, ret=0, r=i;
			for(int j=1;j<=n;j++) b[j]=a[j]%p, r-=a[j]/p;
			sort(b+1, b+n+1, comp);
			for(int j=1;j<=r;j++) ret+=p-b[j];
			if(ret<=k) {
				cout<<p;
				return 0;
			}
		}
	}
	for(int i=sqrt(sum+0.5);i>=1;i--){
		if(sum%i==0){
			int p=i, now=0, ret=0, r=i;
			for(int j=1;j<=n;j++) b[j]=a[j]%p, r-=a[j]/p;
			sort(b+1, b+n+1, comp);
			for(int j=1;j<=r;j++) ret+=p-b[j];
			if(ret<=k) {
				cout<<p;
				return 0;
			}
		}
	}
}