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>
#define rep(i,n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;


int main()
{
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  int s = 0;
  rep(i, n)
  {
    cin >> a[i];
    s += a[i];
  }
  vector<int> c;
  rep(i, s)
    if (s % (i + 1) == 0)
      c.push_back(i + 1);
  reverse(c.begin(), c.end());
  // rep(i, c.size())
  //   cout << c[i] << endl;
  for (int i : c)
  {
    vector<int> m(n);
    int msum = 0;
    rep(j, n)
    {
      m[j] = a[j] % i;
      msum += m[j];
    }
    sort(m.begin(), m.end());
    int msum2 = 0;
    for (int j = m.size() - 1; j >= 0; j--)
    {
      msum -= m[j];
      msum2 += i - m[j];
      if (max(msum, msum2) <= k)
      {
        cout << i << endl;
        return 0;
      }
    }
  }
  return 0;
}