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;
    
#define sz(x) (int) (x).size()
#define INF 0x3f3f3f3f
#define pb push_back
    
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, pii> piii;
    
const int N = 100005;
const int mod = 1000000007;

vector<int> grafo[N];
int memo[N];
bool vis[N];

int solve(int x) {
    if (sz(grafo[x]) == 0) return 0;
    if (memo[x] != -1) return memo[x];
    int ans = 0;
    for (int v: grafo[x]) {
        ans = max(ans, 1 + solve(v));
    }
    return memo[x] = ans;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int n, m, a, b;
    int ans = 0;
    cin >> n >> m;
    while (m--) {
        cin >> a >> b;
        grafo[a].pb(b);
    }
    memset(memo, -1, sizeof(memo));
    for (int i = 1; i <= n; i++) {
        if (vis[i]) continue;
        vis[i] = true;
        ans = max(ans, solve(i));
    }
    cout << ans << "\n";

    //cerr << "\n" << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
    return 0;
}