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: Accepted)

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::*;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::collections::*;

pub fn divisors(n: u64) -> Vec<u64> {
    let mut res = Vec::new();
    let mut x = 1;
    while x * x <= n {
        if n % x == 0 {
            res.push(x);
            let y = n / x;
            if y != x {
                res.push(y);
            }
        }
        x += 1;
    }
    res
}

fn solve() {
    input! {
        n: usize, k: u64,
        av: [u64; n]
    };

    let sum = av.iter().sum::<u64>();
    let mut divisors = divisors(sum);
    divisors.sort_by_key(|&x| -(x as i64));
    for d in divisors {
        let mut dv = av.iter().map(|x| x % d).collect::<Vec<_>>();
        dv.sort();
        let mcusum = dv
            .iter()
            .scan(0, |acc, &x| {
                *acc += x;
                Some(*acc)
            })
            .collect::<Vec<_>>();
        let pcusum = dv
            .iter()
            .scan(0, |acc, &x| {
                *acc += d - x;
                Some(*acc)
            })
            .collect::<Vec<_>>();
        for i in 0..n {
            let msum = mcusum[i];
            let psum = pcusum[n - 1] - pcusum[i];
            let c = std::cmp::max(msum, psum);
            let diff = (msum as i64 - psum as i64).abs() as u64;
            if diff % d == 0 && c <= k {
                println!("{}", d);
                return;
            }
        }
    }
}

fn main() {
    std::thread::Builder::new()
        .name("big stack size".into())
        .stack_size(256 * 1024 * 1024)
        .spawn(|| {
            solve();
        })
        .unwrap()
        .join()
        .unwrap();
}