Given an integer `n`

, return *the number of strings of length *`n`

* that consist only of vowels (*`a`

*, *`e`

*, *`i`

*, *`o`

*, *`u`

*) and are lexicographically sorted.*

A string `s`

is **lexicographically sorted** if for all valid `i`

, `s[i]`

is the same as or comes before `s[i+1]`

in the alphabet.

**Example 1:**

Input:n = 1Output:5Explanation:The 5 sorted strings that consist of vowels only are`["a","e","i","o","u"].`

**Example 2:**

Input:n = 2Output:15Explanation:The 15 sorted strings that consist of vowels only are ["aa","ae","ai","ao","au","ee","ei","eo","eu","ii","io","iu","oo","ou","uu"]. Note that "ea" is not a valid string since 'e' comes after 'a' in the alphabet.

**Example 3:**

Input:n = 33Output:66045

**Constraints:**

`1 <= n <= 50`

```
struct Solution;
impl Solution {
fn count_vowel_strings(n: i32) -> i32 {
let n = n as usize;
let mut res = 0;
for i in 0..5 {
res += Self::dp(i, n);
}
res
}
fn dp(i: i32, n: usize) -> i32 {
if n == 1 {
1
} else {
let mut res = 0;
for j in 0..=i {
res += Self::dp(j, n - 1);
}
res
}
}
}
#[test]
fn test() {
let n = 1;
let res = 5;
assert_eq!(Solution::count_vowel_strings(n), res);
let n = 2;
let res = 15;
assert_eq!(Solution::count_vowel_strings(n), res);
let n = 33;
let res = 66045;
assert_eq!(Solution::count_vowel_strings(n), res);
}
```