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)

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

int calc_change_times(vector<int> numbers, int divisor);

int main(void)
{
    int array_length;
    int max_change_count;
    vector<int> numbers;

    cin >> array_length >> max_change_count;

    int sum = 0;
    for(int i = 0; i < array_length; i++){
        int n;
        cin >> n;
        sum += n;
        numbers.push_back(n);
    }

    int gcd = 1;
    int div = 1;
    while(div * div <= sum){
        if(sum % div == 0){
            int change_times = calc_change_times(numbers, div);
            if(change_times != -1 && change_times <= max_change_count){
                gcd = max(gcd, div);
            }
            
            change_times = calc_change_times(numbers, sum / div);
            if(change_times != -1 && change_times <= max_change_count){
                gcd = max(gcd, sum / div);
            }
        }

        div += 1;
    }

    cout << gcd << endl;

    return 0;
}

int calc_change_times(vector<int> numbers, int divisor)
{
    int left_sum = 0;
    int right_sum = 0;

    vector<int> exceeds;

    for(int i = 0; i < numbers.size(); i++){
        left_sum += numbers[i] % divisor;
        exceeds.push_back(numbers[i] % divisor);
    }

    sort(exceeds.begin(), exceeds.end());

    int border = numbers.size() - 1;
    while(left_sum != right_sum){
        left_sum -= exceeds[border];
        right_sum += divisor - exceeds[border];

        if(border == 0){
            return -1;
        }

        border -= 1;
    }

    return left_sum; // same as the times to change
}