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>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;

const ll INF64 = 1LL << 60;
const int INF32 = 1 << 29;
const int MOD =   1000000007;

/// Lib
vector<long long> calc_divisor(long long n) {
    vector<long long> res;
    for (long long i = 1LL; i*i <= n; ++i) {
        if (n % i == 0) {
            res.push_back(i);
            long long j = n / i;
            if (j != i) res.push_back(j);
        }
    }
    sort(res.begin(), res.end());
    return res;
}
///
bool check(ll key, const vector<ll>& vec, ll maxtry){
    vector<ll> distance(vec.size());
    for(ll i = 0; i < vec.size(); ++i){
        distance[i] = vec[i] % key;
    }
    sort(distance.begin(), distance.end());

    ll cost = 0;
    ll last = vec.size()-1;
    for(ll i = 0; i < vec.size(); ++i){
        if(distance[i] == 0) continue;
        cost += distance[i];
        distance[last] += distance[i];

        bool flag = true;
        while(flag){
            if(distance[last] >= key){
                ll res = distance[last] - key;
                distance[last] = 0;
                --last;
                distance[last] += res;
            }
            else{
                flag = false;
            }
        }
    }

    if(cost <= maxtry){
        return true;
    }
    return false;
}

int main(){
#ifdef MYLOCAL
    ifstream in("input.txt");
    cin.rdbuf(in.rdbuf());
#endif
    ll n, k; cin >> n >> k;
    vector<ll> a(n);
    for(ll i = 0; i < n; ++i){
        cin >> a[i];
    }
    ll sum = 0;
    for(ll i = 0; i < n; ++i){
        sum += a[i];
    }
    vector<ll> div = calc_divisor(sum);
    ll ans = 0;
    for(auto it : div){
        if(check(it, a, k)){
            ans = max(ans, it);
        }
    }
    cout << ans << endl;
    return 0;
}