Suppose you have `n`

integers labeled `1`

through `n`

. A permutation of those `n`

integers `perm`

(**1-indexed**) is considered a **beautiful arrangement** if for every `i`

(`1 <= i <= n`

), **either** of the following is true:

`perm[i]`

is divisible by`i`

.`i`

is divisible by`perm[i]`

.

Given an integer `n`

, return *the number of the beautiful arrangements that you can construct*.

**Example 1:**

Input:n = 2Output:2Explanation:The first beautiful arrangement is [1,2]: - perm[1] = 1 is divisible by i = 1 - perm[2] = 2 is divisible by i = 2 The second beautiful arrangement is [2,1]: - perm[1] = 2 is divisible by i = 1 - i = 2 is divisible by perm[2] = 1

**Example 2:**

Input:n = 1Output:1

**Constraints:**

`1 <= n <= 15`

```
struct Solution;
impl Solution {
fn count_arrangement(n: i32) -> i32 {
let mut res = 0;
let mut nums = (1..=n).collect();
let n = n as usize;
Self::dfs(0, &mut nums, &mut res, n);
res
}
fn dfs(start: usize, nums: &mut Vec<i32>, all: &mut i32, n: usize) {
if start == n {
*all += 1;
} else {
for i in start..n {
if nums[i] % (start as i32 + 1) == 0 || (start as i32 + 1) % nums[i] == 0 {
nums.swap(start, i);
Self::dfs(start + 1, nums, all, n);
nums.swap(start, i);
}
}
}
}
}
#[test]
fn test() {
let n = 2;
let res = 2;
assert_eq!(Solution::count_arrangement(n), res);
}
```