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 <bits/stdc++.h>
using namespace std;

int main() {
    cin.tie(nullptr); ios::sync_with_stdio(false);

    long N, K; cin>>N>>K;
    vector<int> A(N); for (auto& a : A) cin>>a;

    for (long n = 1e6; n >= 1; n--) {
        vector<long> mods(N);
        for (int i = 0; i < N; i++) {
            mods[i] = A[i]%n;
        }
        sort(mods.begin(), mods.end());
        long cap = K;
        int l = 0, r = N-1;
        while (cap > 0 and l < r) {
            if (mods[l] <= n-mods[r]) {
                cap -= mods[l];
                mods[r] += mods[l];
                mods[r] %= n;
                if (mods[r] == 0)
                    r--;
                mods[l] = 0;
                l++;
            } else {
                cap -= (n-mods[r]);
                mods[l] -= n-mods[r];
                mods[r] = 0;
                r--;
            }
        }

        if (cap >= 0 and mods[l] == 0) {
            cout << n << endl;
            return 0;
        }
    }
    cout << 1 << endl;

    return 0;
}