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: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define int long long
typedef long long ll;
const ll mod=1e9+7;
vector<ll> yaku,array;
void set_yaku(ll n){
  vector<ll> kv;
  for(ll i=1;i*i<=n;i++){
    if(n%i==0) kv.push_back(i);
  }
  for(int i=0;i<kv.size();i++){
    yaku.push_back(kv[i]);
    if(n/kv[i]!=kv[i]) yaku.push_back(n/kv[i]);
  }
  sort(yaku.begin(),yaku.end());
}
bool judge(ll n,ll k){
  priority_queue<ll> pque;
  ll sum=0;
  for(int i=0;i<array.size();i++){
    ll val=(n-(array[i]%n))%n;
    pque.push(val);
    sum+=val;
  }
  for(;sum>0;sum-=n){
    ll max_v=pque.top();
    pque.pop();
    max_v-=n;
    pque.push(max_v-n);
  }
  ll s2=0;
  while(!pque.empty()){
    ll kn=pque.top();
    pque.pop();
    if(kn>0) s2+=kn;
  }
  return s2<=k;
}
signed main(){
  ll n,k,sum=0;
  cin>>n>>k;
  for(int i=0;i<n;i++){
    ll a;
    cin>>a;
    sum+=a;
    array.push_back(a);
  }
  set_yaku(sum);

  int lb=0,rb=yaku.size();
  while(rb-lb>1){
    int mid=(lb+rb)/2;
    ll kans=yaku[mid];
    if(judge(kans,k)) lb=mid;
    else rb=mid;
  }
  cout<<yaku[lb]<<endl;
}