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 <iostream>
#include <vector>
#include <string>
#include <numeric>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <queue>

#define rep(i,n) for(int i=0;i<n;i++)
#define dup(x,y) (((x)+(y)-1)/(y)) // dup * y >= x なる最小のdup.
using namespace std;
typedef long long ll;

vector<ll> divisor(ll n) {
	vector<ll> ret;
	for(ll i=1; i*i<=n; i++) {
		if(n%i==0) {
			ret.push_back(i);
			if(i*i != n) ret.push_back(n/i);
		}
	}
	sort(begin(ret), end(ret));
	return ret;
}


int main() {
	ll N, K;
	cin >> N >> K;
	vector<ll> A(N);
	ll sum = 0;
	rep(i, N) {
		cin >> A[i];
		sum += A[i];
	}

	// 候補 Aの和の約数
	vector<ll> divs = divisor(sum);
	vector<ll> q(N);
	ll max_p = 1;
	for(int i=1; i<divs.size(); i++) {
		ll p = divs[i];
		rep(i, N) {
			q[i] = A[i]%p;
		}
		sort(q.begin(), q.end());

		ll move = 0;
		ll q_sum = 0;
		ll id;
		for(id=0; id<N; id++) {
			move += q[id];
			if(move>K) {
				move -= q[id];
				id--;
				break;
			}
		}
		ll rest = 0;
		id++;
		for(;id<N; id++) {
			rest += p - q[id];
		}
		//cout << p << ": " << move << ' ' << rest << endl;
		if(move>=rest) {
			if(p>max_p) max_p = p;
		}
	}
	
	cout << max_p << endl;

  return 0;
}