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 <vector>
#include <algorithm>
#include <utility>
#include <tuple>
#include <cstdint>
#include <cstdio>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <deque>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <cctype>
#include <functional>
#include <ctime>
#include <cmath>
#include <limits>
#include <numeric>
#include <type_traits>
#include <iomanip>
#include <float.h>
#include <math.h>
using namespace std;
using ll = long long;
unsigned euclidean_gcd(unsigned a, unsigned b) {
    if (a < b) return euclidean_gcd(b, a);
    unsigned r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

ll ll_gcd(ll a, ll b) {
    if (a < b) return ll_gcd(b, a);
    ll r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

class UnionFind {
public:
    vector <ll> par;
    vector <ll> siz;
    UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) {
        for (ll i = 0; i < sz_; ++i) par[i] = i;
    }
    void init(ll sz_) {
        par.resize(sz_);
        siz.assign(sz_, 1LL);
        for (ll i = 0; i < sz_; ++i) par[i] = i;
    }
    ll root(ll x) {
        while (par[x] != x) {
            x = par[x] = par[par[x]];
        }
        return x;
    }
    bool merge(ll x, ll y) {
        x = root(x);
        y = root(y);
        if (x == y) return false;
        if (siz[x] < siz[y]) swap(x, y);
        siz[x] += siz[y];
        par[y] = x;
        return true;
    }

    bool issame(ll x, ll y) {
        return root(x) == root(y);
    }

    ll size(ll x) {
        return siz[root(x)];
    }
};

long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

long long modinv(long long a, long long mod) {
    return modpow(a, mod - 2, mod);
}

vector<int> tpsort(vector<vector<int>>& G) {
    int V = G.size();
    vector<int> sorted_vertices;
    queue<int> que;
    vector<int> indegree(V);
    for (int i = 0; i < V; i++) {
        for (int j = 0; j < G[i].size(); j++) {
            indegree[G[i][j]]++;
        }
    }
    for (int i = 0; i < V; i++) {
        if (indegree[i] == 0) {
            que.push(i);
        }
    }
    while (que.empty() == false) {
        int v = que.front();
        que.pop();
        for (int i = 0; i < G[v].size(); i++) {
            int u = G[v][i];
            indegree[u] -= 1;
            if (indegree[u] == 0) que.push(u);
        }
        sorted_vertices.push_back(v);
    }
    return sorted_vertices;
}

int main() {
    ll n, m;
    cin >> n >> m;
    vector<vector<int>> z(n);
    for (int i = 0; i < m; i++) {
        ll a, b;
        cin >> a >> b;
        a--;
        b--;
        z[a].push_back(b);
    }
    vector<int> sorted = tpsort(z);
    vector<int> dp(n);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < z[sorted[i]].size(); j++) {
            dp[z[sorted[i]][j]] = max(dp[z[sorted[i]][j]], dp[sorted[i]] + 1);
        }
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
        ans = max(ans, dp[i]);
    }
    cout << ans << endl;
}