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 Rust (1.15.1) (judge verdict: Wrong Answer)

// Practice AtCoder
// author: Leonardone @ NEETSDKASU

fn main() {
    let mut stdin = String::new();
    std::io::Read::read_to_string(
        &mut std::io::stdin(),
        &mut stdin).unwrap();
    let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get {
        ($t:ty) => (get().parse::<$t>().unwrap());
        () => (get!(i64));
    }
    
    let n = get!(usize);
    let k = get!();
    let mut vs = vec![];
    let mut sum = 0;
    for i in 0 .. n {
        vs.push(get!());
        sum += vs[i];
    }
    let mut ans = 1;
    let mut x = 1;
    while x*x <= sum {
        if sum%x==0 {
            let mut c = vec![0; n];
            let mut t = 0;
            for i in 0 .. n {
                c[i] = vs[i] % x;
                t += c[i];
            }
            c.sort();
            let mut p = 0;
            let mut q = n-1;
            let mut u = 0;
            while p < q {
                c[q] += c[p];
                u += c[p];
                c[p] = 0;
                p += 1;
                if c[q] >= x {
                    c[q-1] += c[q] % x;
                    q -= 1;
                }
            }
            if c[q] == 0 && (u <= k||t-u <= k) {
                ans = std::cmp::max(ans, x);
            }
        } else {
            x += 1;
            continue;
        }
        let y = sum / x;
        if sum%y==0 {
            let x = y;
            let mut c = vec![0; n];
            let mut t = 0;
            for i in 0 .. n {
                c[i] = vs[i] % x;
                t += c[i];
            }
            c.sort();
            let mut p = 0;
            let mut q = n-1;
            let mut u = 0;
            while p < q {
                c[q] += c[p];
                u += c[p];
                c[p] = 0;
                p += 1;
                if c[q] >= x {
                    c[q-1] += c[q] % x;
                    q -= 1;
                }
            }
            if c[q] == 0 && (u <= k||t-u <= k) {
                ans = std::cmp::max(ans, x);
            }
        }
        x += 1;
    }
    println!("{}", ans);
}