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;
#define REP(i, init, n) for(int i = (int)(init); i < (int)(n); i++)

#define vi vector<int>
#define vl vector<long>
#define vvi vector<vector<int>>
#define vvl vector<vector<long>>
#define pint pair<int, int>
#define plong pair<long, long>

vi A;

long solve_move(int mod, int N){
    vi A_MOD(N);    
    int move = 0;
    long mod_sum = 0;
    REP(i, 0, N) {
        A_MOD[i] = A[i] % mod;
        mod_sum += A_MOD[i];
    }
    
    sort(A_MOD.rbegin(), A_MOD.rend());
    REP(i, 0, mod_sum/mod){
        move += mod - A_MOD[i];
       // cout<<mod<<" "<<A_MOD[i]<<endl;
    }
    return move;
}

int main() {
    int N, K;
    cin>>N>>K;
    long sum = 0;
    A.resize(N);
    REP(i, 0, N){
        cin >> A[i];
        sum += A[i];
    }
    int max_GCD = 0;
    int root_sum = pow(sum, 0.5);
    REP(i, 1, root_sum + 1){
        if(sum % i == 0){
            int move = solve_move(i, N);

            if(move <= K && max_GCD < i) max_GCD = i;
            //cout<<move<<" "<<i<<endl;
            move = solve_move(sum/i, N);
            if(move <= K && max_GCD < sum / i) max_GCD = sum / i;
             //cout<<move<<" "<<i<<endl;
        }
    }
    cout << max_GCD << endl;
}