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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <utility>
#include <stack>
using namespace std;
#define rep(i, n) for(int i = 0; i < (n); i ++)
using ll = long long;
const int INF = 1<<25;
typedef pair<int, int> P;
int N, M;
int xs[100000], ys[100000];
vector<int> G[100000];
vector<int> Ginv[100000];
vector<int> sorted;
int degree[100000];
int dp[100000];

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);
    cin >> N >> M;
    rep(i, M) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        G[x].push_back(y);
        Ginv[y].push_back(x);
        degree[y] ++;
    }
    stack<int> st;
    rep(i, N) if (degree[i] == 0) st.push(i);
    while (st.size() > 0) {
        int i = st.top(); st.pop();
        sorted.push_back(i);
        for (int j : G[i]) {
            degree[j] --;
            if (degree[j] == 0) st.push(j);
        }
    }
    rep(i, N) {
        int x = sorted[i];
        int d = 0;
        for (int y : Ginv[x]) {
            d = max(d, dp[y] + 1);
        }
        dp[x] = d;
    }
    int ans = 0;
    rep(i, N) ans = max(ans, dp[i]);
    cout << ans << endl;
    return 0;
}