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: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<math.h>
#include<string>
using namespace std;

long long int a[1000000];
int n;
int hoge(int che){
	long long int d[100000];
	for(int i=0;i<n;i++){
		d[i]=a[i]%che;
	}
	sort(d,d+n);
	//	for(long long int i=0;i<n;i++){cout<<d[i]<<endl;}
	int bid=-1;int lid=n;
	int b=0;int l=0;
	while(bid!=lid-1){
		if(l>=b){bid++;b+=d[bid];}
		else{lid--;l+=(che-d[lid]);}
	}
	return l;
}


int main(){
	long long int sum=0;
	cin>>n;
	int k;cin>>k;
	for(int i=0;i<n;i++){
		cin>>a[i];
		sum+=a[i];
	}
	long long int div[100000];
	int id=-1;
	for(long long int i=1;i<=sqrt(sum);i++){
		if(sum%i==0){
			id++;
			div[id]=i;
			if(i*i!=sum){id++;div[id]=sum/i;}
		}
	}
	sort(div,div+id+1);

	long long int OK=0;
	long long int NG=id+1;
	while(NG!=OK+1){
		long long int check=(NG+OK)/2;
		//cout<<"W";
		int t=hoge(div[check]);
		//cout<<div[check]<<" "<<t<<endl;
		if(t>k){NG=check;}
		else{OK=check;}
	}
	cout<<div[OK]<<endl;
}