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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author alireza_kaviani
 */

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

template<class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;

#define all(x)                      (x).begin(),(x).end()
#define Sort(x)                     sort(all((x)))
#define X                           first
#define Y                           second
#define Mp                          make_pair
#define sep                         ' '
#define endl                        '\n'
#define debug(x)                    cerr << #x << " = " << x << endl
#define SZ(x)                       ll(x.size())
#define fast_io                     ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define set_random                  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

ll poww(ll a, ll b, ll md) {
    return (!b ? 1 : (b & 1 ? a * poww(a * a % md, b / 2, md) % md : poww(a * a % md, b / 2, md) % md));
}

const ll MAXN = 500 + 10;
const ll LOG = 22;
const ll INF = 8e18;
const ll MOD = 1e9 + 7; // 998244353; // 1e9 + 9;

ll n, k, A[MAXN], B[MAXN], sum, ans;

void calc(ll x) {
    set<pll> st;
    for (ll i = 1; i <= n; i++) B[i] = A[i] % x, st.insert({B[i], i});
    ll cost = 0;
    while (st.size() != 0) {
        pll p = *st.begin();
        ll t = p.X;
        st.erase(st.begin());

        while (t) {
            pll q = *prev(st.end());
            st.erase(q);
            ll mn = min(t, x - q.X);
            q.X = (q.X + mn) % x;
            t -= mn;
            cost += mn;
            st.insert(q);
        }
    }
    if (cost <= k) ans = max(ans, x);
}

int main() {
    fast_io;

    cin >> n >> k;
    for (ll i = 1; i <= n; i++) cin >> A[i], sum += A[i];

    for (ll i = 1; i * i <= sum; i++) {
        if (sum % i == 0) {
            calc(i);
            calc(sum / i);
        }
    }
    cout << ans << endl;

    return 0;
}
/*

*/