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 (1.42.0) (judge verdict: Wrong Answer)

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let k: i32 = s.next().unwrap().parse().unwrap();
	let a: Vec<i32> = s.take(n).map(|a| a.parse().unwrap()).collect();
	let sum: i32 = a.iter().sum();
	let ans = divisors(sum).into_iter().fold(1, |r, x| {
		let mut sum = 0;
		let mut a: Vec<i32> = a.iter().map(|a| a % x).inspect(|a| sum += a).collect();
		let m = (sum / x) as usize;
		a.sort();
		let cnt: i32 = a[0..n - m].iter().sum();
		if cnt <= k {
			x
		} else {
			r
		}
	});
	println!("{}", ans);
}

pub fn divisors<T>(n: T) -> Vec<T>
where
	T: num::PrimInt,
{
	let n = if n < T::zero() { T::zero() - n } else { n };
	let mut d = T::one();
	let mut divs = vec![];
	while d * d <= n {
		if n % d == T::zero() {
			divs.push(d);
			if d != n / d {
				divs.push(n / d);
			}
		}
		d = d + T::one();
	}
	divs
}