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)

//! ----------------------------------------------
//! Framework <https://github.com/vain0x/procon>
//!
//! See the bottom of file for solution.
//! ----------------------------------------------

#![allow(unused_imports)]
#![allow(non_snake_case)]

use std::cell::RefCell;
use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::fmt::{Debug, Display, Formatter, Write as FmtWrite};
use std::io::{stderr, stdin, BufRead, Write};
use std::mem::{replace, swap};
use std::ops::*;
use std::rc::Rc;

/// Print values to standard error if debug mode.
#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), stderr());
            writeln!(err, "\x1B[33m{}\x1B[0m = {:?}", e, $e).unwrap()
        })*
    };
}

/// Read from standard input and parse each word.
/// - `read!(T, U, ..)` parses a line as a tuple of words.
/// - `read![[T]]` parses a line as an array of words.
/// - `read![..; N]` parses `N` lines, using `read!(..)` repeatedly.
#[allow(unused_macros)]
macro_rules! read {
    ([$t:ty] ; $n:expr) =>
        ((0..$n).map(|_| read!([$t])).collect::<Vec<_>>());
    ($($t:ty),+ ; $n:expr) =>
        ((0..$n).map(|_| read!($($t),+)).collect::<Vec<_>>());
    ([$t:ty]) =>
        (rl().split_whitespace().map(|w| w.parse().unwrap()).collect::<Vec<$t>>());
    ($($t:ty),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(w.next().unwrap().parse::<$t>().unwrap()),*)
    }};
}

/// Read a line from standard input.
#[allow(dead_code)]
fn rl() -> String {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();

    #[allow(deprecated)]
    buf.trim_right().to_owned()
}

// -----------------------------------------------
// Solution
// -----------------------------------------------

fn divisors(x: i64) -> Vec<i64> {
    let mut d = 1;
    let mut ds = vec![];
    while d * d <= x {
        if x % d == 0 {
            ds.push(d);
            ds.push(x / d);
        }
        d += 1
    }
    ds.sort();
    ds.dedup();
    ds
}

// x を d で割った余り
fn rep(x: i64, d: i64) -> i64 {
    x % d
}

// x に足すと d の倍数になる最小値
fn complement(x: i64, d: i64) -> i64 {
    (d - x % d) % d
}

fn solve(A: &mut [i64], K: i64) -> i64 {
    let N = A.len();
    let sum = A.iter().sum::<i64>();

    for d in divisors(sum).into_iter().rev() {
        // d で割った余りだけ考える。
        // 余りの小さい値を -1 して、大きい値を +1 するため、余りに関して昇順にする。
        A.sort_by(|&l, &r| rep(l, d).cmp(&(rep(r, d))));

        // A[..i] を減少させる量
        let mut neg = 0;
        // A[i..] を増加させる量
        let mut pos = A.iter().map(|&a| complement(a, d)).sum::<i64>();

        for i in 0..N + 1 {
            // 減少量と増加量が一致して操作回数がどちらも K 以下なら操作が存在する
            if neg <= K && pos <= K && neg == pos {
                return d;
            }

            if i < N {
                neg += A[i] % d;
                pos -= complement(A[i], d);
            }
        }
    }

    1
}

fn main() {
    let (_, K) = read!(usize, i64);
    let mut A = read![[i64]];
    println!("{}", solve(&mut A, K))
}