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 <string>
#include <utility>
#include <algorithm>
#include <cmath>
using namespace std;
long long int n, k, a[500], b[500], suma = 0, sumb = 0, w;
int main(){
	long long int count = 0;
	cin >> n >> k;
	for(long long int i = 0;i < n;i++){
		cin >> a[i];
		suma += a[i];
	}
	long long int i2;
	long long int s = sqrt(suma);
	for(long long int i = 1;i <= s;i++){
		sumb = 0;
		if((suma % i) == 0){
			i2 = suma / i;
			for(long long int j = 0;j < n;j++){
				b[j] = a[j] % i2;
				sumb += b[j];
			}
			sort(b, b + n);
			w = sumb / i2;
			count = 0;
			for(long long int q = 0;q < w;q++){
				count += (i2 - b[n - q - 1]);
			}
			if(count <= k){
				cout << i2 << endl;
				return 0;
			}
		}
	}
	for(long long int i = s;i > 0;i--){
		sumb = 0;
		if((suma % i) == 0){
			for(long long int j = 0;j < n;j++){
				b[j] = a[j] % i;
				sumb += b[j];
			}
			sort(b, b + n);
			w = sumb / i;
			count = 0;
			for(long long int q = 0;q < w;q++){
				count += (i - b[n - q - 1]);
			}
			if(count <= k){
				cout << i << endl;
				return 0;
			}
		}
	}
}