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)

/*
(ai, m - ai)

m = 7
2 5 3 0 6 5

5 2     6
    4 7   2
    
0 2 3 5 5 6

    
max(\sum ai, km - \sum aj)


*/

#include <cstdio>
#include <algorithm>

typedef long long int64;

const int MAXN = 2048 + 10;
const int64 INF = 0x3fffffffffffffff;

int n, k, S;
int a[MAXN];
int d[MAXN], dn;

int64 judge(int m) {
	static int s[MAXN];
	int i, x;
	int64 ans = INF;
	
	for(i=1; i<=n; i++)
		s[i] = a[i] % m;
	std::sort(s+1, s+1+n);
	for(i=1; i<=n; i++) s[i]+=s[i-1];
	for(i=0; i<=n; i++) 
		ans = std::min(ans, std::max(1ll * s[i], 1ll * m * (n - i) - s[n] + s[i]));
	return ans;
}

int main() {
	int i, ans = 0;
	
	scanf("%d %d", &n, &k);
	for(i=1; i<=n; i++) {
		scanf("%d", &a[i]);
		S+=a[i];
	}
	
	for(i=1; i*i<=S; i++) {
		if(!(S%i)) {
			dn++;
			d[dn] = i;
			if(i*i<S) {
				dn++;
				d[dn] = S / i;
			}
		}
	}
	
	for(i=1; i<=dn; i++)
		if(ans < d[i] and judge(d[i]) <= k)
			ans = d[i];
	printf("%d\n", ans);
	return 0;
}