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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define REP(i,a) for(int i = 0; i < (a); i++)
#define ALL(a) (a).begin(),(a).end()
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
const long long LINF = 1e18;
const long long MOD = 1e9 + 7;

int ans = 0;
vector<int> dp;

vector<int> Tsort(vector<vector<int>> & G){
    vector<int> To; //トポロジカルソート後のグラフ
    // To.size() == V であればトポロジカルソートができた、そうでなければできなかったと判定できる
    int V = G.size();
    stack<int> st; //入次数が 0 の頂点の集合
    int indeg[V] = {}; //頂点の入次数を管理する配列
    for(int i = 0; i < V; i++){
        for(int j = 0; j < G[i].size(); j++){
            indeg[G[i][j]]++;
        }
    }
    for(int i = 0; i < V; i++){
        if(indeg[i] == 0){
            st.push(i);
        }
    }
    dp.assign(V, 0);
    while(!st.empty()){
        int u = st.top();
        st.pop();
        To.push_back(u);
        for(int i = 0; i < G[u].size(); i++){
            indeg[G[u][i]]--; //隣接する頂点の入次数を1減らす
            dp[G[u][i]] = max(dp[G[u][i]], dp[u] + 1);
            ans = max(ans, dp[G[u][i]]);
            if(indeg[G[u][i]] == 0){
                st.push(G[u][i]); //入次数が 0 になればstに追加
            }
        }
    }
    return To;
}

signed main(){
    int n,m;
    cin >> n >> m;
    vector<vector<int>> G(n);
    int x,y;
    REP(i,m){
        cin >> x >> y;
        x--;
        y--;
        G[x].push_back(y);
    }
    Tsort(G);
    cout << ans << endl;
    return 0;
}