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 ll long long
#define ull unsigned long long
#define si(X) scanf("%d", &(X))
#define sll(X) scanf("%lld",&(X))
#define INFL 0x3f3f3f3f3f3f3f3fLL
const ll mod = 1e9 + 7;
const int M = 2e6 + 5;
ll gcd(ll a,ll b){if(b==0)return a;return gcd(b,a%b);}
ll expo(ll base,ll pow){ll ans = 1;
   if(pow < 0) return 0;
   while(pow!=0){if(pow&1==1){ans = ans*base;ans = ans;}
   base *= base;pow/=2;}return ans;}
ll expo1(ll base,ll pow){ll ans = 1;
   if(pow < 0) return 0;
   while(pow!=0){if(pow&1==1){ans = ans*base;ans = ans%mod;}
   base *= base;base%=mod;pow/=2;}return ans;}   
ll inv(ll x){return expo1(x,mod-2);}
ll mult_mod(ll a , ll b){a = a%mod; b = b%mod;ll c = (a * b)%mod;return c;}
ll add_mod(ll a , ll b){a = (a + mod)%mod;b = (b + mod)%mod;ll c = (a + b)%mod;return c;}
ll fac[(int)(2e6 + 7)] , inv_fac[(int)(2e6 + 7)];
void precomp_factorial(){fac[0] = 1;for(ll i = 1 ; i <= M ; i++) fac[i] = mult_mod(i , fac[i - 1]);
    inv_fac[M] = inv(fac[M]);for(ll i = M - 1 ; i >= 0 ; i--) inv_fac[i] = mult_mod(i + 1 , inv_fac[i + 1]);}
ll nCk(ll n , ll k){if(k == 0) return 1LL;
    ll ans = fac[n];ans = mult_mod(ans , inv_fac[n - k]);ans = mult_mod(ans , inv_fac[k]);return ans;}
int dx[8] = {1 , 0 , -1 , 0 , 1 , -1 , -1 , 1};    // last 4 diagonal
int dy[8] = {0 , 1 , 0 , -1 , 1 , 1 , -1 , -1};

int dp[123456],vis[123456];
vector <vector <int> > vec;
int ans = 0;
void dfs(int n){
     dp[n] = 1;
     vis[n] = 1;
     int max1 = 0;
     for(auto x:vec[n]){
         if(!vis[x]){
            dfs(x);
         }
         max1 = max(max1,dp[x]);
     }
     dp[n] += max1;
     ans = max(ans, dp[n]);
}

int main(){
    int n, m;
    cin >> n >> m;
    vec.resize(n+1);
    for(int i = 0; i < m; i++){
       int u,v;
       cin >> u >> v;
      vec[u].push_back(v);
    }
    for(int i = 1; i <= n; i++){
        if(!vis[i]) dfs(i);
    }
    cout << ans-1;
}