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: Accepted)

macro_rules! get {
    (Vec<$t:ty>) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.split_whitespace()
                .map(|t| t.parse::<$t>().unwrap())
                .collect::<Vec<_>>()
        }
    };
    ($t:ty) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t:ty),*) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            (
                $(iter.next().unwrap().parse::<$t>().unwrap(),)*
            )
        }
    };
    ($t:ty; $n:expr) => {
        (0..$n).map(|_|
            get!($t)
        ).collect::<Vec<_>>()
    };
    ($($t:ty),*; $n:expr) => {
        (0..$n).map(|_|
            get!($($t),*)
        ).collect::<Vec<_>>()
    };
}

use std::cmp;

fn f(i: i64, N: usize, A: Vec<i64>) -> i64 {
    let mut B = A.iter().map(|a| a % i).collect::<Vec<_>>();
    B.sort();
    let mut k = N - 1;
    let mut q = 0;
    for j in 0..N {
        while B[j] > 0 && B[j] < i {
            if j == k {
                return 10000000000;
            }
            let d = cmp::min(i - B[k], B[j]);
            B[k] += d;
            B[j] -= d;
            if B[k] == i {
                k -= 1;
            }
            q += d;
        }
    }
    q
}

fn main() {
    let (N, K) = get!(usize, i64);
    let A = get!(Vec<i64>);

    // 解の候補はsum(A)の約数である
    let s = A.iter().sum::<i64>();
    let D = (1..s).take_while(|d| d * d <= s).collect::<Vec<_>>();
    let D2 = (1..s).take_while(|d| d * d <= s).map(|d| s / d).collect::<Vec<_>>();
    let ans = D.iter().chain(D2.iter())
        .filter(
            |&d|
            {
                let mut B = A.iter()
                    .map(|a| a % d)
                    .collect::<Vec<_>>();
                B.sort();
                let mut acc = vec!(0; N + 1);
                let mut rev_acc = vec!(0; N + 2);
                for i in 0..N {
                    acc[i + 1] = B[i] + acc[i];
                    rev_acc[N - i] = (d - B[N - i - 1]) + rev_acc[N - i + 1]
                }
                (0..(N + 1)).any(|i| acc[i] <= K && acc[i] == rev_acc[i + 1])
            }
        )
        .max()
        .unwrap();

    println!("{}", ans);
}