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<algorithm>
#include<vector>
#include<cstring>

#define ll long long

using namespace std;

const int NMAX = 1e5+5;

ll v[NMAX],N,K,dif[NMAX];

bool check(int MOD)
{

}


int main()
{
    ll i;

    cin>>N>>K;
    ll sum = 0;
    for(i = 1 ; i <= N ; ++i)
    {
        cin>>v[i];
        sum += v[i];
    }
    ll maxim = 1;
    for(i = 1 ; i*i <= sum ; ++i)
    {
        if(sum % i == 0)
        {
            if(check(i))
                maxim = max(maxim , i);
            bool ok;
            ll MOD = i;
            ll sum1 = 0, ii,j;
            ll cnt = 0;
            for(ii = 1 ; ii <= N ; ++ii)
            {
                int v1 = v[ii]%MOD , v2 = (-v[ii])%MOD;
                if(abs(v1) > abs(v2))
                    dif[ii] = v2 , cnt += abs(v2);
                else
                    dif[ii] = v1 , cnt += abs(v1);
                sum1 += dif[ii];
            }
            ok = (sum1 % MOD == 0) && (cnt/2 <= K);

            if(ok)
                maxim = max(maxim , i);

            MOD = sum/i;
            sum1 = 0, ii,j;
            cnt = 0;
            for(ii = 1 ; ii <= N ; ++ii)
            {
                int v1 = v[ii]%MOD , v2 = (-v[ii])%MOD;
                if(abs(v1) > abs(v2))
                    dif[ii] = v2 , cnt += abs(v2);
                else
                    dif[ii] = v1 , cnt += abs(v1);
                sum1 += dif[ii];
            }
            ok = (sum1 % MOD == 0) && (cnt/2 <= K);
            if(ok)
                maxim = max(maxim , sum/i);
        }
    }

    cout<<maxim;
}