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 <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <tuple>
#include <vector>

using namespace std;

uint64_t a[500];
vector<uint64_t> factors;
uint64_t r[500];

int main()
{
    int n;
    uint64_t k;

    cin >> n >> k;

    uint64_t asum = 0;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        asum += a[i];
    }

    for (uint64_t i = 1; i * i <= asum; i++)
    {
        if (asum % i == 0)
        {
            factors.push_back(i);
            if (i * i != asum)
            {
                factors.push_back(asum / i);
            }
        }
    }

    uint64_t maxFactor = 0;
    for (const uint64_t f : factors)
    {
        for (int i = 0; i < n; i++)
        {
            r[i] = a[i] % f;
        }
        sort(r, r + n);

        int i = 0;
        int j = n - 1;
        uint64_t cnt = 0;
        while (i < j)
        {
            uint64_t p = r[i];
            cnt += p;
            r[i] = 0;

            while (p > 0)
            {
                uint64_t q = min(p, f - r[j]);
                r[j] += q;
                p -= q;
                if (f == r[j])
                {
                    j--;
                }
            }
            i++;
        }
        if (r[j] == 0 && cnt <= k)
        {
            maxFactor = max(maxFactor, f);
        }
    }

    cout << maxFactor << endl;
    return 0;
}