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 <stdio.h>
#include <algorithm>
int n, k, s, m, t, ans, x;
int a[500], b[500];
int main() {
	scanf("%d%d", &n, &k);
	for (int i = 0; i < n; i++) {
		scanf("%d", a + i);
		s += a[i];
	}
	for (int i = 1; i < 23000; i++) {
		if (s % i == 0) {
			m = s / i;
			t = 0;
			for (int j = 0; j < n; j++) {
				b[j] = a[j] % m;
				t += b[j];
			}
			std::sort(b, b + n);
			t /= m;
			t = n - t;
			x = 0;
			for (int j = 0; j < t; j++) {
				x += b[j];
			}
			if (x <= k)ans = m;
		}
		if (ans) {
			printf("%d\n", ans);
			return 0;
		}
	}
	for (int i = 23000; i > 0; i--) {
		if (s % i == 0) {
			m = i;
			t = 0;
			for (int j = 0; j < n; j++) {
				b[j] = a[j] % m;
				t += b[j];
			}
			std::sort(b, b + n);
			t /= m;
			t = n - t;
			x = 0;
			for (int j = 0; j < t; j++) {
				x += b[j];
			}
			if (x <= k)ans = m;
		}
		if (ans) {
			printf("%d\n", ans);
			return 0;
		}
	}
}