Problem:
Score : 100 points

Problem Statement

There is a directed graph G with N vertices and M edges.
The vertices are numbered 1, 2, \ldots, N, and for each i (1 \leq i \leq M), the i-th directed edge goes from Vertex x_i to y_i.
G does not contain directed cycles.

Find the length of the longest directed path in G.
Here, the length of a directed path is the number of edges in it.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq x_i, y_i \leq N

All pairs (x_i, y_i) are distinct.

G does not contain directed cycles.

Input

Input is given from Standard Input in the following format:

N M
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print the length of the longest directed path in G.

Sample Input 1

4 5
1 2
1 3
3 2
2 4
3 4

Sample Output 1

3

The red directed path in the following figure is the longest:

Sample Input 2

6 3
2 3
4 5
5 6

Sample Output 2

2

The red directed path in the following figure is the longest:

Sample Input 3

5 8
5 3
2 3
2 4
5 2
5 1
1 4
4 3
1 3

Sample Output 3

3

The red directed path in the following figure is one of the longest:

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
const int nax = 1e5 + 5;
vector<int> edges[nax];
int in_degree[nax]; // the number of edges going to 'b'
int dist[nax];
bool visited[nax];

void dfs(int a) {
    assert(!visited[a]);
    visited[a] = true;
    for(int b : edges[a]) {
        dist[b] = max(dist[b], dist[a] + 1);
        --in_degree[b];
        if(in_degree[b] == 0) {
            dfs(b);
        }
    }
}

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    while(m--) {
        int a, b;
        scanf("%d%d", &a, &b);
        edges[a].push_back(b);
        ++in_degree[b];
    }
    for(int i = 1; i <= n; ++i) {
        if(!visited[i] && in_degree[i] == 0) {
            dfs(i);
        }
    }
    int answer = 0;
    for(int i = 1; i <= n; ++i) {
        answer = max(answer, dist[i]);
    }
    printf("%d\n", answer);
}