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 <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <utility>
#include <cmath>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <numeric>
#include <functional>
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef pair<ll, ll> P;
#define rep(i, n) for(ll i = 0; i < n; i++)
#define exrep(i, a, b) for(ll i = a; i <= b; i++)
#define out(x) cout << x << endl
#define exout(x) printf("%.10f\n", x)
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define pb push_back
#define re0 return 0
const ll mod = 1000000007;
const ll INF = 1e16;
const ll MAX_N = 100010;

// nの約数を降順に列挙したvectorを返す。計算量はO(√n)
vector<ll> yakusu(ll n) {
    vl res;
    for(ll i = 1; i*i <= n; i++) {
        if(n%i == 0) {
            res.pb(i);
            if(i != n/i) {
                res.pb(n/i);
            }
        }
    }
    sort(rall(res));
    return res;  
}
 
int main() {
    ll n, k;
    cin >> n >> k;

    vl a(n);
    ll sum = 0;
    rep(i, n) {
        cin >> a[i];
        sum += a[i];
    }

    vl AAA = yakusu(sum);

    for(auto d : AAA) {
        vl v(n);
        rep(i, n) {
            v[i] = a[i]%d;
        }
        sort(all(v));
        rep(x, n) {
            ll cnt = 0;
            exrep(i, 0, x) {
                cnt += v[i];
            } 
            if(cnt > k) {
                continue;
            }
            cnt = 0;
            exrep(i, x+1, n-1) {
                cnt += d - v[i];
            }
            if(cnt <= k) {
                out(d);
                re0;
            }
        }
    }
    
    re0;
}