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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;
import java.util.Arrays;
public class Main{
	static int[] a;
	static int n;
	static int k;
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		n=sc.nextInt();
		k=sc.nextInt();
		a=new int[n];
		long sum=0;
		for(int i=0; i<n; i++){
			a[i]=sc.nextInt();
			sum+=a[i];
		}
		long ans=1;
		for(int i=1; i*i<=sum; i++){
			if(sum%i==0){
				ans=Math.max(ans,ch(i));
				ans=Math.max(ans,ch(sum/i));
			}
		}
		System.out.println(ans);
	}
	public static long ch(long t){
		long[] ar=new long[n+1];
		for(int i=1; i<=n; i++){
			ar[i]=a[i-1]%t;
		}
		Arrays.sort(ar);
		for(int i=2; i<=n; i++){
			ar[i]+=ar[i-1];
		}
		int q=(int)(ar[n]/t);
		if(ar[n-q]<=k){
			return t;
		}else{
			return 1;
		}
	}
}