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<bits/stdc++.h>
using namespace std;
long long res = 0, sum = 0, k, a[501], n;
void check (long long val) {
  multiset <long long> s;
  long long sumf = 0, sumz = 0;
  for (int i = 1; i <= n; ++ i) {
    s.insert (a[i] % val);
    sumf += (val - (a[i] % val));
  }
  for (auto v : s) {
    sumz += v;
    sumf -= (val - v);
    if (sumz == sumf && sumz <= k) {
      res = max (res, val);
    }
  }
}
int main () {
  if (fopen ("test.inp", "r")) {
    freopen ("test.inp", "r", stdin);
  }
  ios_base :: sync_with_stdio (0);
  cin.tie (nullptr);
  cout.tie (nullptr);
  cin >> n >> k;
  for (int i = 1; i <= n; ++ i) {
    cin >> a[i];
    sum += a[i];
  }
  for (long long i = 1; i <= sqrt (sum); ++ i) {
    if (sum % i == 0) {
      check (i);
      check (sum / i);
    }
  }
  cout << res;
}