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)

/**
 * @Author:      G_bg
 * @DateTime:    2019-08-04 20:37:41
 * @Description: 
 */
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double PI = 3.1415926535898;
const double E = 2.718281828459;
const int INF = 0x7fffffff;
const int mod = 1e9+7;
const int maxn = 500+10;
int a[maxn],n,k,b[maxn];
bool judge(int x){
    for(int i = 0;i < n;i++) b[i] = a[i]%x;
    sort(b,b+n);
    int cnt = 0;
    int l = 0,r = n-1;
    while(l < r){
        int mi = min(b[l],x-b[r]);
        cnt += mi;
        b[l] -= mi;
        (b[r] += mi)%=x;
        if(b[l] == 0) l++;
        if(b[r] == 0) r--;
    }
    return cnt <= k;
}
int main(int argc, char const *argv[]) {
    ios::sync_with_stdio(0);cin.tie(0);
    cin >> n >> k;
    int sum = 0;
    for(int i = 0;i < n;i++) cin >> a[i],sum += a[i];
    sort(a,a+n);
    int sq = sqrt(sum);
    int ma = 1;
    for(int i = 1;i <= sq;i++){
        if(sum%i == 0){
            if(judge(i) && i > ma) ma = i;
            if(judge(sum/i) && sum/i > ma) ma = sum/i;
        }
    }
    cout << ma << endl;
    //system("pause");
    return 0;
}