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 <cstdio>
#include <algorithm>
#include <cstring>
#include <cassert>
using namespace std;

int n , a[505] , sum , k;

int factor[100005] , tot;

pair<int , int> pa[505];

long long check(int fac)
{
	long long pos = 0 , neg = 0;
	for(int i = 1 ; i <= n ; i ++)
	{
		int low =  a[i] / fac * fac;
		pa[i] = make_pair(a[i] - low , fac - a[i] + low);
		pos += fac - a[i] + low;
	}
	sort(pa + 1 , pa + 1 + n);
	for(int i = 1 ; i <= n ; i ++)
	{
		neg += pa[i].first , pos -= pa[i].second;
		if(neg == pos) return neg;
	}
	assert(0); 
}

int main()
{
	scanf("%d %d" , &n , &k);
	for(int i = 1 ; i <= n ; i ++)
		scanf("%d" , a + i) , sum += a[i];
		
	for(int i = 1 ; i * i <= sum ; i ++)
		if(sum % i == 0)
		{
			factor[++tot] = i;
			if(i * i != sum) factor[++tot] = sum / i;
		}
	sort(factor + 1 , factor + 1 + tot);
	for(int i = tot ; i ; i --)
		if(check(factor[i]) <= k)
			return printf("%d\n" , factor[i]) & 0;
	assert(0);
}