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.io.*;
import java.util.*;

class Main {
    static Scanner scanner = new Scanner();

    public static void main(String[]$) {
        int n = scanner.nextInt();
        int k = scanner.nextInt();
        long[] a = new long[n];
        long sum = 0;
        for (int i = 0; i < n; i++) {
            sum += a[i] = scanner.nextInt();
        }
        Set<Long> divisors = divisors(sum);
        for (long d : divisors) {
            long[] b = new long[n];
            long m = 0;
            for (int i = 0; i < n; i++) {
                m += b[i] = a[i] % d == 0 ? 0 : a[i] / d * d + d - a[i];
            }
            Arrays.sort(b);
            long cost = 0;
            m /= d;
            for (int i = n - 1; i >= 0; i--) {
                if (--m < 0) {
                    break;
                }
                cost += d - b[i];
            }
            if (cost <= k) {
                System.out.println(d);
                return;
            }
        }
    }

    static Set<Long> divisors(long n) {
        Set<Long> divisors = new TreeSet<>(Comparator.reverseOrder());
        for (long i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                divisors.add(i);
                divisors.add(n / i);
            }
        }
        return divisors;
    }

    static class Scanner {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in), 32768);
        StringTokenizer tokenizer;

        String next() {
            try {
                while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                    tokenizer = new StringTokenizer(reader.readLine());
                }
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return tokenizer.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}