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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <set> 
#include <map>
#include <bitset>
#include <cctype>
#include <cmath>
#include <cstring>
#include <iomanip>
#define REP(i,n) for(int i=0; i<(n); ++i)
#define FOR(i,a,b) for(int i=(a); i<(b); ++i)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const int INF = 100100100;
const ll INFLL = 1001001001001001001;  // > 10^18
const int MOD = (int)1e9 + 7;
const ll MODLL = (ll)1e9 + 7;
const double EPS = 1e-9;

vector<ll> enum_divisor (ll n) {
    vector<ll> res;
    for (ll i = 1; i*i <= n; i++) {
        if (n % i == 0) {
            res.push_back(i);
            if (i != (n/i)) {
                res.push_back(n/i); 
            }
        }
    }
    return res;
}

int main () {

    ll n,k; cin >>n>>k;
    vector<ll> a(n); REP(i,n) cin >> a[i];

    ll s = 0;
    for (ll i = 0; i < n; i++) {
        s += a[i];
    }

    vector<ll> d = enum_divisor(s);

    ll mx = 0;
    for (ll i = 0; i < d.size(); i++) {
        vector<ll> b = a;
        ll sum = 0;
        for (ll j = 0; j < n; j++) {
            b[j] %= d[i];
            sum += b[j];
        }
        if (sum % d[i] == 0) {
            sort(b.begin(), b.end(), greater<ll>());
            for (ll l = 0; l < (sum/d[i]); l++) {
                sum -= b[l];
            }
            if (sum <= k) {
                mx = max(mx, d[i]);
            } 
        } 
    }

    cout << mx << endl;
    return 0;
}