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;

const int maxn = 505;

int n, k, sum, ans;
int a[maxn];
vector < int > d;

signed main () {
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n; ++i) {
        scanf("%d", &a[i]);
        sum += a[i];
    }
    for (int i = 1; i * i <= sum; ++i) {
        if (sum % i == 0) {
            d.push_back(i);
            d.push_back(sum / i);
        }
    }
    for (int x: d) {
        vector < int > b;
        int add = 0, rm = 0;
        for (int i = 0; i < n; ++i) {
            if (a[i] % x) {
                b.push_back(a[i] % x);
                add += x - b.back();
            }
        }
        sort(b.begin(), b.end());
        int cur = add;
        for (int i = 0; i < int(b.size()); ++i) {
            rm += b[i];
            add -= x - b[i];
            cur = min(cur, max(rm, add));
        }
        if (cur <= k)
            ans = max(x, ans);
    }
    printf("%d\n", ans);
}