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: Wrong Answer)

/**
    ▂▃▅▇█▓▒░۩۞۩ ۩۞۩░▒▓█▇▅▃▂
        Make it happen now, not tomorrow.
        Tomorrow is a  loser's excuse.
    ▂▃▅▇█▓▒░۩۞۩ ۩۞۩░▒▓█▇▅▃▂
**/
#ifndef _GLIbCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>

#if __cplu >= 201103L
#include <ccomplex>
#include <cfenv>
#include <cinttypes>
#include <cstdalign>
#include <cstdbool>
#include <cstdint>
#include <ctgmath>
#include <cwchar>
#include <cwctype>
#include <cstring>
#endif

// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include<random>
#include <chrono>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>

#if __cplu >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#include <time.h>
#endif
#define __ ios_base::sync_with_stdio(false);cin.tie(NULL);
using namespace std;
// #define INF 0x7f7f7f7f
#define ull unsigned long long int
#define pii pair<int,int>
#define ll long long int
#define ld long double
#define MSET(x,y) memset(x,y,sizeof(x))
// #define l long
#define se second
#define fi first
#define NIL 0
#define UNVISITED -1
#define EXPOLORED 0
#define mk make_pair
#define pb push_back
#define sc(n) scanf("%d",&n)
#define NIL 0
#define pii pair<int,int>
#define eps 1e-9
#define all(x) x.begin(),x.end()
#define me(u,v,w) make_pair(w,make_pair(u,v))
#define sqr(x) x*x
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define inrange(x,y,row,col) (x>=0 && y>=0 && x<row && y<col)
#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++ e)
// #define add(a,b) (a + b >= MOD ? a+b-MOD:a+b)
#define hs(c1,c2) (256*min(c1,c2) + max(c1,c2))
typedef struct{ int sum,suf,pre,max;} Node;
int toint(const string &s) { stringstream ss; ss << s; int x; ss >> x; return x; }
const int MAXN = 3010;
// const int N = 2e5+10;
const int UP = 31;
const int lim = 1e18+1;
const ll mod = 1e9+7;
const ll inf = 3e18*1LL;
// const int inf = 1e9+10;
// const double pi = 3.141592653589793238462643383;
const double pi = acos(-1);
const int INF = 1e9;
const double Phi =  1.618033988749894;
const int logn = 20;
const double phi = 0.618033988749894;
const double root5 = 2.236067977;
/**
    ▂▃▅▇█▓▒░۩۞۩ ۩۞۩░▒▓█▇▅▃▂
        Make it happen now, not tomorrow.
        Tomorrow is a loser's excuse.
    ▂▃▅▇█▓▒░۩۞۩ ۩۞۩░▒▓█▇▅▃▂
There is nothin in a catterpillar that tells you its going to be a butterfly ---------!
//--- Matrix Expo//

//-- Matix Expo//
**/
const int N = 1e5+10;
vector<int> gr[N];
int in[N];
int dp[N];
vector<int> topo;
bool vis[N];
string ans;
void dfs(int x){
    if(vis[x])return;
    vis[x] = 1;
    for(int v:gr[x]){
        dp[v] = max(dp[v],dp[x]+1);
        --in[v];
        if(in[v]==0){
            dfs(v);
        }
    }
}
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;++i){
        int x,y;
        scanf("%d%d",&x,&y);
        gr[x].pb(y);
        ++in[y];
    }
    for(int i=1;i<=n;++i){
        if(!vis[i]){
            dfs(i);
        }
    }
    int ans =0;
    for(int x:dp){
        ans = max(ans,x);
    }
    cout<<ans<<endl;
    return 0;
}