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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		Main mainObj = new Main();
		mainObj.solove();
	}
	
	public void solove() {
		
		Scanner sc = new Scanner(System.in);
		
		String[] firstArr = sc.nextLine().split(" ");
		int n = Integer.parseInt(firstArr[0]);
		int k = Integer.parseInt(firstArr[1]);
		
		int sumOfA = 0;
		int[] aArr = new int[n];
		
		String[] strArr = sc.nextLine().split(" ");
		sc.close();
		
		for(int i = 0 ; i < n;i++) {
			int nextInt = Integer.parseInt(strArr[i]);
			aArr[i] = nextInt;
			sumOfA += nextInt;
		}
		
		ArrayList<Integer> divisorArr = new ArrayList<>();
		for(int i = 1 ; i*i <= sumOfA ; i++) {
			if(sumOfA % i == 0) {
				divisorArr.add(new Integer(i));
				divisorArr.add(new Integer(sumOfA/i));
			}
		}
		Collections.sort(divisorArr);
		
		for(int i = divisorArr.size() - 1; i >= 0 ; i--) {
			if(canDivide(aArr, divisorArr.get(i), k)) {
				System.out.println(divisorArr.get(i));
				return;
			}
		}
	}
	
	public boolean canDivide(int[] aArr, int divideValue, int k) {
		
		int sumOfR = 0;
		boolean hasZero = false;
		ArrayList<Integer> rList = new ArrayList<>();
		for(int i = 0; i < aArr.length; i++) {
			if(aArr[i] % divideValue != 0) {
				rList.add(aArr[i] % divideValue);
				sumOfR += aArr[i] % divideValue;
			}else {
				hasZero = true;
			}
		}
		
		if(sumOfR % divideValue != 0) {
			return false;
		}
		
		if(rList.size() == 0) {
			return true;
		}
		
		if(hasZero && 
				((sumOfR % divideValue == 0 && sumOfR <= k) 
						|| ((divideValue * rList.size() - sumOfR) % divideValue == 0 && (divideValue * rList.size() - sumOfR) <= k))) {
			return true;
		}
		
		Collections.sort(rList);
		
		int cumSum = 0;
		for(int i = 0; i < rList.size() - 1; i++) {
			cumSum += rList.get(i);
			int plusSum = divideValue * (rList.size() - i - 1) -  (sumOfR - cumSum);
			
//			if(cumSum < plusSum) {
//				if((plusSum - cumSum)%divideValue == 0 && plusSum <= k) {
//					return true;
//				}
//			}
//			if(plusSum < cumSum) {
//				if((cumSum - plusSum)%divideValue == 0 && cumSum <= k) {
//					return true;
//				}
//			}
			
			if(plusSum - cumSum == 0 && cumSum <= k && plusSum <= k) {
				return true;
			}
		}
		
		return false;		
	}
}