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.*;

public class Main {
    static final TreeMap<Long, Integer> factorization = new TreeMap<>();
    static final List<Long> factors = new ArrayList<>();

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        long ln = n;
        long k = in.nextInt();
        long[] array = new long[n];
        long sum = 0;
        for (int i = 0; i < n; i++) {
            array[i] = in.nextLong();
            sum += array[i];
        }
        for (long p = 2, sumTemp = sum; p < 22361; p++) {
            int power;
            for (power = 0; sumTemp % p == 0; sumTemp /= p, power++);
            if (power > 0) {
                factorization.put(p, power);
            }
        }
        if (factorization.isEmpty()) {
            factorization.put(sum, 1);
        }
        generateFactors(1);
        factors.sort(Comparator.reverseOrder());
        for (long f : factors) {
            Long[] residues = new Long[n];
            long residueSum = 0;
            for (int i = 0; i < n; i++) {
                residues[i] = (long) array[i] % f;
                residueSum += residues[i];
            }
            Arrays.sort(residues);
            long req = 0;
            long li = 0;
            for (int i = 0; li < ln - (residueSum / f); i++, li++) {
                req += residues[i];
            }
            if (k >= req) {
                System.out.println(f);
                return;
            }
        }
    }

    static void generateFactors(long curr) {
        if (factorization.isEmpty()) {
            factors.add(curr);
        } else {
            long p = factorization.firstKey();
            int power = factorization.remove(p);
            for (int e = 0; e <= power; e++, curr *= p) {
                generateFactors(curr);
            }
            factorization.put(p, power);
        }
    }
}