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: Runtime Error)

#include <bits/stdc++.h>

#define pb push_back
#define INF 1e9
#define mod 998244353
#define mod2 1000000007
#define dbg(x) cout << (#x) << " --> " << x << "\n";
#define dbgarr(y)                                                                                            \
  cout << (#y) << " --> " << '\n';                                                                           \
  for(auto& x : y) cout << x << '\n';
using namespace std;

typedef priority_queue<int, vector<int>, greater<int>> minPq;
typedef priority_queue<int> maxPq;
typedef long long ll;

void goodLuckHaveFun() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  /* clang-format off */
  #ifdef _DEBUG
  freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
  #endif
  /* clang-format on */
}

vector<int> adList[10010];

int dp[100010];

int dfs(int n, int dist){
  for(auto x: adList[n]){
    if(dp[x] > dist+1)continue;
    dp[x] = max(dp[x], dist+1);
    dfs(x,dist+1);
  }
}

int main() {
  goodLuckHaveFun();
  int n, m;
  cin >> n >> m;
  pair<int, int> p;
  
  for(int i = 0; i < m; ++i) {
    cin >> p.first >> p.second;
    adList[p.first].pb(p.second);
  }

  for(int i = 1; i <= n; ++i){
    if(dp[i] == 0)dfs(i,0);
  }
  cout << *max_element(dp,dp+10001) << '\n';
  
}