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: Wrong Answer)

import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.SortedSet;
import java.util.TreeSet;
import java.util.stream.Collectors;

class Main {
    static int[] primes;
    static List<Integer> primeList;
    static {
        int length = (int) Math.sqrt(500000000) + 1;
        primes = new int[length];
        int max = (int) Math.sqrt(length) + 1;
        for (int i = 2; i < max; i++) {
            if (primes[i] == 1) {
                continue;
            }
            int cur = i * i;
            while (cur < length) {
                primes[cur] = 1;
                cur += i;
            }
        }

        primeList = new ArrayList<>();
        for (int i = 2; i < length; i++) {
            if (primes[i] == 0) {
                primeList.add(i);
            }
        }
    }

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        int N = in.nextInt();
        int K = in.nextInt();
        int[] A = new int[N];
        for (int i = 0; i < N; i++) {
            A[i] = in.nextInt();
        }

        int sum = 0;
        for (int a : A) {
            sum += a;
        }

        SortedSet<Integer> factors = new TreeSet<>(Comparator.reverseOrder());
        List<Integer> tmpPrimes = new ArrayList<>(primeList);
        while (sum > 1 && tmpPrimes.size() > 0) {
            if (sum % tmpPrimes.get(0) == 0) {
                sum /= tmpPrimes.get(0);
                factors.addAll(factors.stream().map(i -> i*tmpPrimes.get(0)).collect(Collectors.toList()));
                factors.add(tmpPrimes.get(0));
            } else {
                tmpPrimes.remove(0);
            }
        }
        factors.add(sum);
        boolean hasResult = false;
        for (int i : factors) {
            if (helper(i, K, A)) {
                System.out.println(i);
                hasResult = true;
                break;
            }
        }
        if (!hasResult) {
            System.out.println(1);
        }
    }

    private static boolean helper(int i, int k, int[] A) {
        PriorityQueue<Integer> q1 = new PriorityQueue<>(Comparator.reverseOrder()), q2 = new PriorityQueue<>(Comparator.reverseOrder());
        int sum = 0;
        for (int j = 1; j < A.length; j++) {
            int e = A[j] % i;
            q1.add(e);
            sum += e;
        }

        q2.add(sum);
        while (q1.peek() + q2.peek() > i || (q1.peek() + q2.peek() == i && q1.peek() < q2.peek())) {
            int x = q2.poll();
            int y = q1.poll();
            q2.add(i - y);
            if (x > i) {
                q2.add(x - i);
            } else {
                q1.add(i - x);
            }
            sum -= y + Math.max(0, i - x);
        }

        return sum <= k;
    }
}