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;
using ll=long long;
const ll MOD=1000000007;
const double PI=3.14159265358979;
const ll INF= pow(10,18);
typedef pair<ll,ll> P;
typedef vector<ll> vl;
typedef vector<vl> vvl;
#define FOR(i,a,b) for(ll i=a;i<b;i++)
#define rep(i,n) FOR(i,0,n)
string abc="abcdefghijklmnopqrstuvwxyz";
string ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
struct edge{ll to,cost;};

int main() {
  ll ans=1;
  ll n,k;
  cin >> n >> k;
  vl a(n);
  ll sum=0;
  rep(i,n){
    cin >> a[i];
    sum += a[i];
  }
  ll z=sqrt(sum);
  vl p(100000);
  ll j=0;
  FOR(i,1,z+1){
    if(sum%i==0){
      p[j]=i;
      j++;
    }
  }
  j--;
  ll x;
  if(sum==z*z){
    p.resize(2*j+1);
    FOR(i,j+1,2*j+1){
      p[i]=sum/p[2*j-i];
    }
    x=2*j+1;
  }
  else{
    p.resize(2*j+2);
    FOR(i,j+1,2*j+2){
      p[i]=sum/p[2*j+1-i];
    }
    x=2*j+2;
  }
  vl b(n);
  vl c(n);
  rep(i,x){
    ll MIN=INF;
    rep(j,n){
      b[j]=a[j]%p[i];
    }
    sort(b.begin(),b.end());
    rep(j,n){
      c[j]=(p[i]-b[j])%p[i];
    }
    vl sumb(n);
    vl sumc(n);
    sumb[0]=b[0];
    sumc[n-1]=c[n-1];
    rep(i,n-1){
      sumb[i+1]=sumb[i]+b[i+1];
      sumc[n-2-i]=sumc[n-1-i]+c[n-2-i];
    }
    rep(j,n-1){
      MIN=min(MIN,max(sumb[j],sumc[j+1]));
    }
    if(MIN<=k){
      ans=p[i];
    }
  }
  cout << ans << endl;
}