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)

#define _USE_MATH_DEFINES
#include <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <complex>
#include <string>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <limits>
#include <climits>
#include <cfloat>
#include <functional>
#include <iterator>
#include <memory>
#include <regex>
using namespace std;

void findDivisor(int n, vector<int>& divisor)
{
    divisor.clear();
    stack<int> stk;
    for(int i=1; i*i<=n; ++i){
        if(n % i == 0){
            divisor.push_back(i);
            if(i * i < n)
                stk.push(n / i);
        }
    }
    while(!stk.empty()){
        divisor.push_back(stk.top());
        stk.pop();
    }
}

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

    vector<int> div;
    findDivisor(sum, div);
    reverse(div.begin(), div.end());
    for(int x : div){
        vector<int> v(n);
        for(int i=0; i<n; ++i)
            v[i] = a[i] % x;
        sort(v.rbegin(), v.rend());

        long long cnt1 = accumulate(v.begin(), v.end(), 0LL);
        long long cnt2 = 0;
        bool ok = (cnt1 <= k);
        for(int i=0; i<n; ++i){
            cnt1 -= v[i];
            cnt2 += (x - v[i]) % x;
            ok |= (cnt1 <= k && cnt2 <= k);
        }
        if(ok){
            cout << x << endl;
            break;
        }
    }

    return 0;
}