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<bits/stdc++.h>
using namespace std;
#define ll long long
const ll p=1000000007ll;
const int N=200005;
template<class IT>inline void cmin(IT &a,IT b){if(b<a)a=b;}
template<class IT>inline void cmax(IT &a,IT b){if(a<b)a=b;}
int a[N],b[N],n,u,s;
inline void solve(int x){
//	printf("solve %d\n",x);
	int i,t,s2=0;u=0;
	for(i=1;i<=n;++i)s2+=(b[i]=a[i]%x);
	sort(b+1,b+n+1);
	t=n-s2/x;
	for(i=1;i<=t;++i)u+=b[i];
}
int main(){
	int i,t,k,ans=0;
	scanf("%d%d",&n,&k);
	for(i=1;i<=n;++i){scanf("%d",&a[i]);s+=a[i];}
//	printf("sum=%d\n",s);
	t=sqrt(s);
	for(i=1;i<=t;++i)if(!(s%i)){
		solve(  i);if(u<=k)cmax(ans,  i);
		solve(s/i);if(u<=k)cmax(ans,s/i);
	}
	printf("%d",ans);
	return 0;
}