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 <bits/stdc++.h>

using namespace std;

string s;
int a[555];
int n;
int k;
int at;
map<pair<pair<int, int>, int> , int> dp;

int poss(int pos, int neg, int i) {
    if(pos > k || neg > k ) return false;
    if((i >= n) && (pos == neg)) return true;
    if(i >= n) return false;
    if(dp.find(make_pair(make_pair(pos, neg), i)) != dp.end()) return dp[make_pair(make_pair(pos, neg), i)];
    int rem = a[i] % at;
    int x = poss(pos, neg+rem, i+1);
    int y = poss(pos+at-rem, neg, i+1);
    if(x+y > 0) return dp[make_pair(make_pair(pos, neg), i)] = 1;
    else return dp[make_pair(make_pair(pos, neg), i)] = 0;

}


int main() {
    cin >> n >> k;
    int tot = 0;
    for(int i = 0; i < n; i++) {
        cin >> a[i];
        tot+=a[i];
    }

    sort(a, a+n);
    for(int j = 1000000; j >= 1; j--) {
        
        if(tot % j != 0) continue;
        at = j;
        dp.clear();
        if (poss(0, 0, 0)) {
            cout << at;
            return 0;
        }
        
    }


}