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 <map>
#include <set>
#include <string>
#include <queue>
#include <list>
#include <cmath>
#include <stack>
#include <iomanip>

using namespace std;
typedef long long ll;
using Pll = pair<ll,ll>;

const ll INF = 1LL<<60;
const ll MOD = 1000000007;

//cout << fixed << setprecision(10);

const ll N_MAX = 100010;
vector<ll> arr(N_MAX);


void Eratosthenes(){
	for(int i = 2; i < N_MAX; i++){
		arr[i] = 1;
	}
	for(int i = 2; i < sqrt(N_MAX); i++){
		if(arr[i]){
			for(int j = 0; i * (j + 2) < N_MAX; j++){
				arr[i *(j + 2)] = 0;
			}
		}
	}
    //arr[i]=1が素数
}

int main() {
    ll N,K;
    cin >> N >> K;
    vector<ll> A(N);
    for(int i=0;i<N;i++) cin >> A[i];

    sort(A.begin(),A.end());
    ll sum=0;
    for(int i=0;i<N;i++){
        sum+=A[i];
    }
    if((sum-A[N-1])<=K){
        cout << sum << endl;
        return 0;
    }
    /*
    Eratosthenes();
    vector<ll> fact(N_MAX,0);
    bool prime=true;
    ll sumtmp=sum;
    vector<Pll> ps;
    for(int i=2;i<N_MAX;i++){
        if(arr[i]=1){
            while(sumtmp%i==0){
                if(prime)prime=false;
                fact[i]++;
                sumtmp/=i;
            }
            if(fact[i]!=0){
                ps.push_back(pll(i,fact[i]));
            }
        }
    }
    
    //1つ以外0のときは上でやったので素数の時は1しかない
    if(prime){
        cout << 1 << endl;
        return 0;
    }
    */

    vector<ll> nums;
    for(int i=1;i<=sqrt(sum);i++){
        if(sum%i==0){
            nums.push_back(i);
            nums.push_back(sum/i);
        }
    }

    if(nums.size()==0){
        cout << 1 << endl;
        return 0;
    }
    
    sort(nums.rbegin(),nums.rend());

    
    for(int i=0;i<nums.size();i++){
        ll div=nums[i];
        ll sums=0;
        for(int j=0;j<N;j++){
            ll tmp=A[j]/div;
            sums+=min(llabs(A[j]-tmp*div),llabs(A[j]-(tmp+1)*div));
        }
        if(sums/2<K){
            cout << nums[i] << endl;
            return 0;
        }
    }

    cout << 1 << endl;






    return 0;
}