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)

#ifdef OJ_MISC_OFFLINE

#include "prolog.h"

#else
#define LOG
#define PRINT
#define LCERR
#include <iostream>
auto optimize_cpp_stdio = []() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    std::setvbuf(stdout, nullptr, _IOFBF, BUFSIZ);
    return 0;
}();
#endif

#include <algorithm>
#include <array>
#include <cassert>
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <numeric>
#include <string>
#include <vector>

namespace {

using namespace std;

class Solution {
public:
    int max_gcd(const vector<int>& A, int64_t K) {
        const int N = A.size();
        int sum_A = accumulate(A.begin(), A.end(), 0);
        vector<int> cand;
        for (int i = 1; i <= sum_A; ++i) {
            if (sum_A % i == 0) {
                cand.emplace_back(i);
            }
        }
//        int lb = 0;
//        int ub = cand.size();
//        while(lb < ub) {
//            int mid = lb + (ub - lb) / 2;
//            int cur_cand = cand[mid];
//            if (validate(cur_cand, A, K)) {
//                LOG("Candidate {} OK\n", cur_cand);
//                lb = mid + 1;
//            } else {
//                LOG("Candidate {} NG\n", cur_cand);
//                ub = mid;
//            }
//        }
//        return cand[lb - 1];
        int best = -1;
        for (int cur_cand : cand) {
            if (validate(cur_cand, A, K)) {
                best = cur_cand;
            }
        }
        return best;
    }
private:
    bool validate(int cand_gcd, const vector<int>& A, int64_t K) {
        LOG("Validating candidate {}\n", cand_gcd);
        const int N = A.size();
        static vector<int> remainders;
        remainders.clear();
        for (int i = 0; i < N; ++i) {
            int rem = A[i] % cand_gcd;
            if (rem > 0) {
                remainders.emplace_back(rem);
            }
        }
        sort(remainders.begin(), remainders.end());
//        int64_t sum_remainder = accumulate(remainders.begin(), remainders.end(), 0ll);
//        LOG("{}\n", sum_remainder);
//        int64_t t = sum_remainder / cand_gcd;
//        int64_t sum_abs_diff = 0;
//        for (int i = 0; i < remainders.size() - t; ++i) {
//            sum_abs_diff += remainders[i];
//        }
//        for (int i = remainders.size() - t; i < remainders.size(); ++i) {
//            sum_abs_diff += cand_gcd - remainders[i];
//        }
//        if (sum_abs_diff > 2 * K) {
//            return false;
//        } else {
//            return true;
//        }
        int l = 0;
        int r = remainders.size() - 1;
        int cost = 0;
        while(l < r) {
            int diff = min(remainders[l], cand_gcd - remainders[r]);
            cost += diff;
            remainders[l] -= diff;
            remainders[r] += diff;
            if (remainders[l] == 0){
                ++l;
            }
            if (remainders[r] == cand_gcd) {
                --r;
            }
        }
        return cost <= K;
    }
};

int solve() {
    Solution solution;
    // handle IO here
    int N;
    int64_t K;
    cin >> N >> K;
    vector<int> A(N);
    for (int i = 0; i < N; ++i) {
        cin >> A[i];
    }
    cout << solution.max_gcd(A, K) << '\n';
    return 0;
}

} // anonymous namespace

#ifdef OJ_MISC_OFFLINE

#include "epilog.h"

#else

int main() {
    return solve();
}

#endif