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>
#define ALL(v) v.begin(), v.end()
#define MOD 1000000007
#define INF 1000000011
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i++)
using namespace std;
typedef long long ll;
 
int main(){
 ll n,k;
  cin>>n>>k;
  ll val[n+1];
  rep(i, n){
    cin>>val[i+1];}
  ll sum=0;
  rep(i, n){
    sum+=val[i+1];}
  
  vector <ll> vec{};
  for(int i=1;i*i<=sum;i++){
    if(sum%i==0){
      vec.push_back(i);
      vec.push_back(sum/i);
    }
  }
  
  sort(ALL(vec));
  reverse(ALL(vec));
  
  ll ans=0;
  
  
  for(int j=0;j<=vec.size();j++){
  
  ll mod[n+1];
  rep(i, n){
    mod[i+1]=0;}
    
    ll mo=vec[j];
    
    rep(i, n){
      ll an=val[i+1]%mo;
      mod[i+1]=an;
    }
   
    sort(mod, mod+n+1, greater<ll>());
    
    ll wa=0;
    rep(i, n){
      wa+=mod[i+1];}
    ll ks=wa/mo;
    
    wa=0;
    for(int h=ks+1;h<=n;h++){
      wa+=mod[h];}
    
    if(wa<=k){
      ans=mo;
      break;}
    
    
    
    
    
  }
  
  cout<<ans<<endl;
    
  
  
 
}