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 <vector>
#include <algorithm>
#include <numeric>
using namespace std;


bool canDivide(vector<int>& A, int d, int K){
    int N = A.size();
    vector<int> rem(N);  // 余り
    for(int i = 0; i < N; ++i) rem[i] = A[i]%d;
    sort(rem.begin(), rem.end());
    // 累積和
    vector<int> r(rem), l(rem);  // 右からの累積和と左からの累積和
    for(int i = 1; i < N; ++i) r[i] += r[i-1];
    for(int i = 0; i < N; ++i) l[i] = d - l[i];
    for(int i = N-2; i >= 0; --i) l[i] += l[i+1];

    for(int i = 0; i < N-1; ++i){
        if(r[i] == l[i+1] and r[i] <= K) return true;
    }
    return false;
}


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

    int sumA = accumulate(A.begin(), A.end(), 0);

    // sumA の約数について全探索
    int ans = 0;
    for(int i = 1; i*i <= sumA; ++i){
        if(sumA%i != 0) continue;
        if(canDivide(A, i, K)) ans = max(ans, i);
        if(canDivide(A, sumA/i, K)) ans = max(ans, sumA/i);
    }
    cout << ans << endl;

    return 0;
}