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"
#include<cassert>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
const long long mod = 1000000007;
const long long inf = 1ll << 61;
const double eps = 0.000000000001;
typedef long long ll;
struct edge { int to; int cost;};
int a[505];
signed main() {
	int n, k; cin >> n >> k;
	rep(i, n)cin >> a[i];
	int s = 0;
	rep(i, n)s += a[i];
	vector<int>V;
	for (int i = 1; i*i <= s; i++) {
		if (s%i == 0) {
			V.push_back(i);
			if (i*i != s)V.push_back(s / i);
		}
	}
	sort(V.begin(), V.end());
	int ans = 0;
	rep(i, V.size()) {
		int asum = 0;
		vector<int>sv;
		rep(j, n) {
			int x = a[j] % V[i];
			sv.push_back(x);
			asum += x;
		}
		sort(sv.begin(), sv.end());
		asum /= V[i];
//		cout << V[i] << ' ' << asum << endl;
		int aasum = 0;
		rep(j, n - asum)aasum += sv[j];

		if (aasum <= k)ans = max(ans, V[i]);
	}
	cout << ans << endl;
}