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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;
long long a[510],b[510],n,k,s = 0,dp[510][510] = {};
bool ch(long long x){
	long long i,j;
	for(i=0;i<=n;i++){
		for(j=0;j<=n;j++){
			dp[i][j] = 1000000000;
		}
	}
	s = 0;
	for(i=0;i<n;i++){
		b[i] = a[i]%x;
		s += b[i];
	}
	dp[0][0] = 0;
	for(i=1;i<=n;i++){
		for(j=0;j<n;j++){
			dp[i][j] = min(dp[i][j],dp[i - 1][j]);
			dp[i][j + 1] = min(dp[i][j + 1],dp[i - 1][j] + b[i - 1]);
		}
	}
	for(i=0;i<=n;i++){
		if(max(dp[n][i],dp[n][i] + (n - i)*x - s)<=k){
			return true;
		}
	}
	return false;
}
vector<long long> v;
int main(){
	int i;
	cin >> n >> k;
	for(i=0;i<n;i++){
		cin >> a[i];
		s += a[i];
	}
	for(i=1;i*i<=s;i++){
		if(s%i==0){
			v.push_back(i);
			v.push_back(s/i);
		}
	}
	sort(v.begin(),v.end());
	for(i=(int)v.size() - 1;i>=0;i--){
		if(ch(v[i])){
			cout << v[i] << endl;
			return 0;
		}
	}
}