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;
#define rep(i,n) for(int i = 0; i < (n);i++)
#define sz(x) int(x.size())
typedef long long ll;
typedef pair<int,int> P;
vector<long long> divisor(long long n){
    vector<long long> res;
    for(long long i = 1; i*i <= n; i++){
        if(n%i==0){
            res.push_back(i);
            if(n/i != i)res.push_back(n/i);
        }
    }
    return res;
}
int main(){ 
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  ll sum = 0;
  rep(i,n) cin >> a[i], sum += a[i];

  auto d = divisor(sum);
  ll res = 0;
  for (auto p : d) {
    vector<ll> cum(n+1,0);
    vector<int> b(n);
    for (int i = 0; i < n; i++) b[i] = a[i] % p;
    sort(b.begin(), b.end());
    for (int i = 0; i < n; i++) cum[i + 1] = cum[i] + b[i];
    bool ok = false;
    for (int i = 0; i <= n; i++) {
      ll r = p * (n - i) - (cum[n] - cum[i]);
      ll l = cum[i];
      if (max(r, l) <= k) ok = true;
    }
    if (ok) res = max(res, p);
  }
  cout << res << endl;
  return 0;
}