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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int factorize(long long n, long long p[], int m[])
{
	int k;
	long long i;
	for (i = 2, k = 0; i * i <= n; i++) {
		if (n % i != 0) continue;
		p[k] = i;
		for (m[k] = 0; n % p[k] == 0; m[k]++, n /= p[k]);
		k++;
	}
	if (n > 1) {
		p[k] = n;
		m[k++] = 1;
	}
	return k;
}

int main()
{
	int i, N, A[501];
	long long K;
	scanf("%d %lld", &N, &K);
	for (i = 1; i <= N; i++) scanf("%d", &(A[i]));
	
	int j, k, m[100];
	long long sum = 0, p[100];
	for (i = 1; i <= N; i++) sum += A[i];
	k = factorize(sum, p, m);
	
	int l[100] = {1}, max;
	long long ans = 1, tmp = p[0];
	while (l[k] == 0) {
		if (tmp > ans) {
			for (i = 1, sum = 0, max = 0; i <= N; i++) {
				j = abs(A[i] - tmp) % tmp;
				j = (j < tmp - j)? j: tmp - j;
				sum += j;
				if (j > max) max = j;
			}
			if (sum - max <= K) ans = tmp;
		}
		
		for (i = 0; i < k && l[i] == m[i]; i++);
		for (l[i]++, tmp *= p[i--]; i >= 0; l[i--] = 0) {
			for (j = 0; j < l[i]; j++) tmp /= p[i];
		}
	}
	
	printf("%lld\n", ans);
	fflush(stdout);
	return 0;
}