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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
int myabs(int n)
{
	if (n < 0)
		n *= -1;
	return n;
}
int main()
{
	int n, k;
	scanf("%d %d", &n, &k);
	int i;
	int a[502];
	for (i = 0; i < n; i++)
		scanf("%d", &a[i]);
	int sum = 0;
	for (i = 0; i < n; i++)
		sum += a[i];
	int c, x, j;
	int count;
	int b[502];
	int p, q;
	for (i = 1; i <= sqrt(sum); i++)
	{
		if (sum % i == 0)
		{
			x = sum / i;
			for (j = 0; j < n; j++)
				b[j] = a[j] % x;
			for (j = 0; j < n - 1; j++)
			{
				p = b[j];
				if (p > x - b[j])p = x - b[j];
				q = b[j + 1];
				if (q > x - b[j + 1])q = x - b[j + 1];
				if (p > q)
				{
					b[j] ^= b[j + 1];
					b[j + 1] ^= b[j];
					b[j] ^= b[j + 1];
					if (j > 0)
						j -= 2;
				}
			}
			c = 0;
			count = 0;
			for (j = 0; j < n - 1; j++)
			{
				if (myabs(c + b[j]) < myabs(c - x + b[j]))
				{
					c += b[j];
					count += myabs(c);
				}
				else
				{
					c -= x - b[j];
					count += myabs(c);
				}
			}
			if (count <= k)
			{
				printf("%d\n", x);
				return 0;
			}
		}
	}
	for (i = (int)sqrt(sum); i > 0; i--)
	{
		if (sum % i == 0)
		{
			x = i;
			for (j = 0; j < n; j++)
				b[j] = a[j] % x;
			for (j = 0; j < n - 1; j++)
			{
				p = b[j];
				if (p > x - b[j])p = x - b[j];
				q = b[j + 1];
				if (q > x - b[j + 1])q = x - b[j + 1];
				if (p > q)
				{
					b[j] ^= b[j + 1];
					b[j + 1] ^= b[j];
					b[j] ^= b[j + 1];
					if (j > 0)
						j -= 2;
				}
			}
			c = 0;
			count = 0;
			for (j = 0; j < n - 1; j++)
			{
				if (myabs(c + b[j]) < myabs(c - x + b[j]))
				{
					c += b[j];
					count += myabs(c);
				}
				else
				{
					c -= x - b[j];
					count += myabs(c);
				}
			}
			if (count <= k)
			{
				printf("%d\n", x);
				return 0;
			}
		}
	}
}