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;
typedef pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second
 
#define bit(n,k) ((n>>k)&1) //*n no k bit me 1 or 0*/

const int mx=100010;
const ll mod=1e9+7;


int main(){  
  ll n,k;
  cin >> n >> k;
  vector<ll> a(n);
  ll sum = 0;
  rep(i,n){
    cin >> a[i];
    sum += a[i];
  }
  priority_queue<ll> q;
  for(int i=1; i<ceil(sqrt(sum)); i++)if(sum%i==0){
    q.push(i);
    q.push(sum/i);
  }
  q.push(0); q.push(sum);

  ll ans = 1;
  while(!q.empty()){
    ll p = q.top(); q.pop();
    ll cnt = longinf;
    vector<ll> calc(n);
    rep(i,n) calc[i] = a[i]%p;
    sort(calc.begin(),calc.end());

    ll pl = 0;
    rep(i,n) pl += p - calc[i];
    ll mi = 0;
    rep(i,n){
      mi += calc[i];
      pl -= p-calc[i];
      cnt = min(cnt, max(pl,mi));
    }
    
    if(cnt<=k){
      ans = p;
      break;
    }
  }
  cout << ans << endl;
  return 0;
}