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)

// https://atcoder.jp/contests/abc136/tasks/abc136_e
//
#![allow(unused_imports)]
use std::io::*;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_macros)]
macro_rules! dvec {
    ($t:expr ; $len:expr) => {
        vec![$t; $len]
    };

    ($t:expr ; $len:expr, $($rest:expr),*) => {
        vec![dvec!($t; $($rest),*); $len]
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        println!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

fn isok(a: &Vec<i32>, d: i32, k: i32) -> bool {
    let n = a.len();
    let mut modulo = vec![];
    for ai in a {
        if ai % d >= 1 {
            modulo.push(ai%d);
        }
    }
    modulo.sort();

    let n = modulo.len();
    let mut sum = vec![0; n+1];
    for i in 0..n {
        sum[i+1] = sum[i] + modulo[i];
    }

    let mut best = k+1;
    for w in 1..n {
        let mut dw = sum[w];
        let mut up = ((n-w) as i32) * d - (sum[n] - sum[w]);
        if (max(up, dw) - min(up, dw)) % d == 0 {
           best = min(best, max(up, dw));
        }
    }
    best <= k
}

fn main() {
    input! {
        n: usize, k: i32,
        a: [i32; n]
    };
    let sum = a.iter().fold(0, |a,c| a+c);

    let mut divisors = vec![];
    for w in 1..sum+1 {
        if w * w >= sum {
            if w * w == sum {
                divisors.push(w);
            }
            break;
        }
        if sum % w == 0 {
            divisors.push(sum/w);
            divisors.push(w);
        }
    }
    divisors.sort();
    divisors.reverse();


    let mut ans = 0;
    for d in divisors {
        if isok(&a, d, k) {
            ans = d;
            break;
        }
    }

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