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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <algorithm>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <functional>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int dx[4] = {-1, 0, 0, 1};
const int dy[4] = {0, -1, 1, 0};

// Self settings
// clang-format off
#define MAX_N 100000
#define MAX_M 100000
#define REP(i, N) for (int i = 0; i < (int)(N); ++i)
// clang-format on
int N, M;
int dp[MAX_N + 1];
vector<int> G[MAX_N + 1];
queue<int> que;
int S[MAX_N];

void solve()
{
  int ans = 0;
  // 始点を追加する
  for (int i = 0; i < N; i++) {
    if (S[i] == 0) que.push(i);
  }

  bool visited[N];
  fill(visited, visited + N, false);
  while (!que.empty()) {
    int c = que.front();
    que.pop();
    queue<P> q;
    for (int i = 0; i < G[c].size(); i++) q.push(P(c, G[c][i]));
    while (!q.empty()) {
      P p = q.front();
      q.pop();
      dp[p.second] = max(dp[p.second], dp[p.first] + 1);
      if (visited[p.second] && dp[p.second] > dp[p.first] + 1) continue;
      for (int i = 0; i < G[p.second].size(); i++)
        q.push(P(p.second, G[p.second][i]));
    }
  }
  for (int i = 0; i < N + 1; i++) ans = max(ans, dp[i]);
  cout << ans << endl;
}

int main(void)
{
  cin >> N >> M;
  REP(i, M)
  {
    int x, y;
    cin >> x >> y;
    x--, y--;
    G[x].push_back(y);
    // 入次 0 のものがかならず最長パスに含まれている
    // (最長パスの途中からのパスは絶対に最長パスではない)
    // ループもないから入次が0の点がかならず存在する
    S[y]++;
  }
  solve();
  return 0;
}