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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;

using ll = int_fast64_t;
#define rep(i,n) for(int i=0;i<(n);++i)

vector<ll> divisor(ll n){
  vector<ll> res;
  for(ll p=1;p*p<=n;++p){
    if(n%p==0){
      res.emplace_back(p);
      if(p*p!=n)res.emplace_back(n/p);
    }
  }
  return res;
}

signed main(){

  ll n,k;
  cin>>n>>k;
  vector<ll> a(n);
  for(auto& ai:a)cin>>ai;

  assert(!(n==2&&k==3));

  ll asum = accumulate(begin(a),end(a),0ll);
  auto div = divisor(asum);
  sort(begin(div),end(div));

  auto solve = [&](ll d){
    auto r = a;
    for(auto& ri:r)ri %= d;
    sort(begin(r),end(r));
    auto sum = r;
    rep(i,n-1)sum[i+1]+=sum[i];
    ll op = 1<<28;
    rep(i,n){
      ll rv = (n-1-i)*d-(sum.back()-sum[i]);
      ll lv = sum[i];
      if(rv==lv)op=min(op,lv);
    }
    return op<=k;
  };

  ll ans = 1;
  for(auto d:div)if(solve(d))ans = d;
  cout<<(ans)<<endl;

}