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<string>
#include<iomanip>
#include<cmath>
#include<vector>
#include<algorithm>

using namespace std;

#define int long long
#define endl "\n"

const long long INF = (long long)1e18;
const long long MOD = (long long)1e9 + 7; 

string yn(bool f){return f?"Yes":"No";}
string YN(bool f){return f?"YES":"NO";}

int K;
const int MAX = 1100000;

int check(vector<int> &A, int x){
	vector<int> div, sum1, sum2;
	int res = INF;
	
	for(int i = 0; i < A.size(); i++){
		if(A[i]%x) div.push_back(A[i]%x);
	}
	
	if(!div.size()) return false;
	
	sort(div.begin(), div.end());
	
	sum1.resize(div.size()+2,0);
	sum2.resize(div.size()+2,0);
	
	for(int i = 0; i < div.size(); i++){
		sum1[i+1] += div[i];
		sum1[i+1] += sum1[i];
	}
	for(int i = div.size(); i >= 1; i--){
		sum2 [i-1] += x-div[i-1];
		sum2[i-1] += sum2[i];
	}
	// cout<<"x = "<<x<<" ------- "<<endl;
	// for(int i = 0; i < div.size(); i++) cout<<div[i]<<" ";cout<<endl;
	// for(int i = 0; i <= div.size(); i++) cout<<sum1[i]<<" ";cout<<endl;
	// for(int i = 0; i <= div.size(); i++) cout<<sum2[i]<<" ";cout<<endl;
	
	for(int i = 0; i <= div.size(); i++){
		// if(sum2[i]-sum1[i] == 0)
			res = min(res, max(sum1[i],sum2[i]));
		
	}
	
	return res <= K;
}

signed main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout<<fixed<<setprecision(10);
	
	int ans = 1;
	int sum = 0;
	
	int N;
	vector<int> A;
	
	cin>>N>>K;
	
	A.resize(N);
	
	for(int i = 0; i < N; i++){
		cin>>A[i];
		sum += A[i];
	}
	
	for(int i = 1; i*i <= MAX*N; i++){
		if(sum%i) continue;
		if(check(A,i)) ans = max(i, ans);
		if(check(A,sum/i)) ans = max(sum/i, ans);
	}
	
	cout<<ans<<endl;
	
	return 0;
}