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 modulo 1000000007
#define mod(mod_x) ((((long long)mod_x))%modulo)
#define Inf 100000000000000000



int main(){
	int N;
	cin>>N;
	long long K;
	cin>>K;
	long long sum = 0;
	vector<long long> A(N);
	for(int i=0;i<N;i++){
		cin>>A[i];
		sum+=A[i];
	}
	
	vector<long long> yaku(0);
	
	for(long long i=1;i*i<=sum;i++){
		if(sum%i==0){
			yaku.push_back(i);
			if(i*i!=sum)yaku.push_back(sum/i);
		}
	}
	
	sort(yaku.begin(),yaku.end(),greater<long long>());
	
	
	for(int i=0;i<yaku.size();i++){
		vector<pair<long long,long long>> B(N);
		long long mi = 0;
		long long pl = 0;
		for(int j=0;j<A.size();j++){
			long long x = (A[j]/yaku[i])*yaku[i];
			long long y = x+yaku[i];
			if(x==A[j])y=x;
			
			B[j].first = -abs(x-A[j]);
			B[j].second = abs(y-A[j]);
			
			mi += B[j].first;
		}
		
		sort(B.begin(),B.end(),greater<pair<long long,long long>>());
		
		
		long long cnt = Inf;
		for(int j=N-1;j>=0;j--){
			if(mi==-pl)cnt = min(cnt,pl);
			mi-=B[j].first;
			pl+=B[j].second;
		}
		
		if(cnt<=K){
			cout<<yaku[i]<<endl;
			return 0;
		}
	}
			
	
	
	return 0;
}