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>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

int N;
int A[505];
int K;

int main() {
    ios_base::sync_with_stdio(false);
    cin >> N >> K;
    for(int i = 1; i <= N; i++)
        cin >> A[i];

    int sum = 0;
    for(int i = 1; i <= N; i++)
        sum += A[i];
    int ans = 1;
    for(int x = sum; x >= 1; x--) {
        if(sum % x != 0)
            continue;
        vector <int> rests;
        for(int i = 1; i <= N; i++) {
            rests.push_back(A[i] % x);
        }
        sort(rests.begin(), rests.end());
        int balance = 0;
        int r = N - 1;
        int rest = 0;
        for(int i = 0; i < N && i <= r; i++) {
            while(i <= r && balance > 0) {
                rest += x - rests[r];
                balance -= (x - rests[r--]);
            }
            if(i > r)
                break;
            rest += rests[i];
            balance += rests[i];
        }

        if(rest <= K * 2) {
            ans = x;
            break;
        }
    }
    cout << ans << endl;
}