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;

int main()
{
    int N, K;
    cin >> N >> K;
    vector<int> numbers(N);
    int sum = 0;
    for(int i = 0; i < N; i++)
    {
        int a;
        cin >> a;
        numbers.at(i) = a;
        sum += a;
    }
    int i = 1;
    vector<int>dcs(0);
    while(i * i < sum)
    {
        if(sum % i == 0)
        {
            dcs.push_back(i);
            dcs.push_back(sum / i);
        }
        i++;
    }
    sort(dcs.begin(), dcs.end(), greater<int>());
    for(int dc: dcs){
        vector<int> mods(0);
        for(int j = 0; j < N; j++){
            if(numbers.at(j) % dc != 0){
                mods.push_back(numbers.at(j) % dc);
            }
        }
        sort(mods.begin(), mods.end());
        vector<int> ruisekiMods(0);
        int ruiseki = 0;
        for(int md: mods){
            ruiseki += md;
            ruisekiMods.push_back(ruiseki);
        }
        for(int j = 0; j < ruisekiMods.size(); j++){
            long long decrementTotal;
            long long incrementTotal;
            decrementTotal = ruisekiMods[j];
            incrementTotal = 1LL * (ruisekiMods.size() - 1 - j) * dc - (ruisekiMods[ruisekiMods.size() - 1] - ruisekiMods[j]);
            if(max(decrementTotal, incrementTotal) <= K){
                cout << dc << endl;
                return 0;
            }
        }
    }
    return 0;
}