You are given an integer `n`

which indicates that we have `n`

courses, labeled from `1`

to `n`

. You are also given an array `relations`

where `relations[i] = [a, b]`

, representing a prerequisite relationship between course `a`

and course `b`

: course `a`

has to be studied before course `b`

.

In one semester, you can study any number of courses as long as you have studied all the prerequisites for the course you are studying.

Return *the minimum number of semesters needed to study all courses*. If there is no way to study all the courses, return `-1`

.

**Example 1:**

Input:n = 3, relations = [[1,3],[2,3]]Output:2Explanation:In the first semester, courses 1 and 2 are studied. In the second semester, course 3 is studied.

**Example 2:**

Input:n = 3, relations = [[1,2],[2,3],[3,1]]Output:-1Explanation:No course can be studied because they depend on each other.

**Constraints:**

`1 <= n <= 5000`

`1 <= relations.length <= 5000`

`1 <= a, b <= n`

`a != b`

- All the pairs
`[a, b]`

are**unique**.

```
struct Solution;
use std::collections::VecDeque;
impl Solution {
fn minimum_semesters(n: i32, relations: Vec<Vec<i32>>) -> i32 {
let n = n as usize;
let mut adj = vec![vec![]; n];
let mut degree = vec![0; n];
for edge in relations {
let x = edge[0] as usize - 1;
let y = edge[1] as usize - 1;
adj[x].push(y);
degree[y] += 1;
}
let mut visited = vec![false; n];
let mut queue = VecDeque::new();
for i in 0..n {
if degree[i] == 0 {
visited[i] = true;
queue.push_back(i);
}
}
let mut res = 0;
while !queue.is_empty() {
let n = queue.len();
res += 1;
for _ in 0..n {
let u = queue.pop_front().unwrap();
for &v in adj[u].iter() {
degree[v] -= 1;
if !visited[v] && degree[v] == 0 {
visited[v] = true;
queue.push_back(v);
}
}
}
}
if visited.into_iter().all(|x| x) {
res
} else {
-1
}
}
}
#[test]
fn test() {
let n = 3;
let relations = vec_vec_i32![[1, 3], [2, 3]];
let res = 2;
assert_eq!(Solution::minimum_semesters(n, relations), res);
let n = 3;
let relations = vec_vec_i32![[1, 2], [2, 3], [3, 1]];
let res = -1;
assert_eq!(Solution::minimum_semesters(n, relations), res);
}
```