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)

#define _CRT_SECURE_NO_WARNINGS
#include "bits/stdc++.h"

#define PB push_back
#define PF push_front
#define LB lower_bound
#define UB upper_bound
#define fr(x) freopen(x,"r",stdin)
#define fw(x) freopen(x,"w",stdout)
#define iout(x) printf("%d\n",x)
#define lout(x) printf("%lld\n",x)
#define REP(x,l,u) for(ll x = l;x<u;x++)
#define RREP(x,l,u) for(ll x = l;x>=u;x--)
#define complete_unique(a) a.erase(unique(a.begin(),a.end()),a.end())
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) a.begin(),a.end()
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define MP make_pair
#define sqr(x) ((x)*(x))
#define lowbit(x) ((x)&(-(x)))
#define lson (ind<<1)
#define rson (ind<<1|1)
#define se second
#define fi first
#define sz(x) ((int)x.size())
#define EX0 exit(0);

typedef  long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
using namespace std;
typedef vector<ll> VLL;
typedef vector<int> VI;
const int block_size = 320;
typedef complex<ll> point;
const ll mod = 1e9 + 7;
const ll inf = 1e9 + 7;
const ld eps = 1e-9;
const db PI = atan(1) * 4;

template<typename T, typename S>inline bool upmin(T&a, const S&b) { return a > b ? a = b, 1 : 0; }
template<typename T, typename S>inline bool upmax(T&a, const S&b) { return a < b ? a = b, 1 : 0; }

ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); }
namespace SOLVE {
	vector<VI> e;
	VI incount;
	void main() {
		int n, m, a, b;
		cin >> n >> m;
		e.resize(n + 5);
		incount.resize(n + 5);

		REP(i, 0, m) {
			cin >> a >> b;
			e[a].push_back(b);
			incount[b]++;
		}

		queue<int> q;
		REP(i, 1, n + 1) {
			if (incount[i] == 0) {
				q.push(i);
			}
		}

		VI maxdistance(n + 5);
		int thisv;
		int ans = 0;
		while (sz(q)) {
			int thisv = q.front();
			for (int v : e[thisv]) {
				maxdistance[v] = max(maxdistance[v], maxdistance[thisv] + 1);
				ans = max(maxdistance[v], ans);
				incount[v]--;
				if (incount[v] == 0) {
					q.push(v);
				}
			}
			q.pop();
		}
		cout << ans;
	}
}


signed main() {

	int t;
	//    in(t);
	t = 1;
	while (t--) {
		SOLVE::main();

	}

	return 0;
}