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: Accepted)

#include<bits/stdc++.h>
#define LL long long
using namespace std;
int n,k,q,Cnt;
LL Sum;
int A[505],C[505];
LL B[5005];
bool Check(LL x) {
	LL need=1e16,w=0;
	for(int i=1; i<=n; ++i)
		C[i]=A[i]%x,w+=C[i];
	sort(C+1,C+1+n);
	LL Pre=C[1];
	for(int i=1; i<=n; ++i,Pre+=C[i])
		if(0==x*(n-i)-w)break;
	return Pre<=k;
}
int main() {
	scanf("%d%d",&n,&k);
	for(int i=1; i<=n; ++i)
		scanf("%d",A+i),Sum+=A[i];
	sort(A+1,A+1+n);
	for(int i=1; 1LL*i*i<=Sum; ++i)
		if(Sum%i==0)B[++Cnt]=i,B[++Cnt]=Sum/i;
	sort(B+1,B+1+Cnt);
	for(int i=Cnt; i>=1; --i)
		if(Check(B[i]))return !printf("%lld",B[i]);
}