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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#include<math.h>
#include<list>
#include <algorithm>
#include<set>
#include<vector>
#include<math.h>
#include<map>
#include<string>
#include <numeric>
#include <queue>
#include <sstream>
#include <bitset>
#include<stack>

using namespace std;
using ll = long long;
using vll = vector<long long>;
using sll = set<long long>;
const long long ll_max = 9223372036854775807;
const long long ll_min = -9223372036854775807;
const int int_max = 2147483647;

template<typename T>
T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<typename T>
T lcm(T a, T b) { return a / gcd(a, b) * b; }

template<typename T>
map<T, T> getPrimeFactor(T n) {
	map<T, T> res;
	for (T i = 2; i * i <= n; ++i) {
		while (n % i == 0) {
			res[i] ++;
			n /= i;
		}
	}
	if (n != 1) res[n] = 1;
	return res;
}

template<typename T>
bool IsPrimeNumber(T num)
{
	if (num <= 2) return true;
	else if (num % 2 == 0) return false;

	double sqrtNum = sqrt(num);
	for (int i = 3; i <= sqrtNum; i += 2)
	{
		if (num % i == 0)
		{
			return false;
		}
	}

	return true;
}

long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}

// 繰り返し二乗法
ll pow2(ll n, ll p, ll mod) {
	if (p == 0 || n == 1) {
		return 1;
	}
	else {
		ll ret = pow2(n * n % mod, p / 2, mod);
		if (p % 2 == 1) {
			ret *= n;
		}
		return ret % mod;
	}
}


#define rep(i,s,e) for(ll i=s;i<e;i++)
#define repeq(i,s,e) for(ll i=s;i<=e;i++)


map<ll,vll> G;
vll t;

ll func(ll pos) {
	ll ret = t[pos];

	if(ret > -1) {
		return ret;
	}

	for (auto x : G[pos]) {
		ret=max(ret,func(x));
	}
	ret++;
	t[pos] = ret;
	return ret;
}


int main() {
	ll N, M;
	cin >> N >> M;
	t.resize(N, -1);
	rep(i, 0, M) {
		ll x, y;
		cin >> x >> y;
		x--; y--;
		G[x].push_back(y);
	}

	ll ans = 0;
	rep(i, 0, N) {
		ans = max(ans, func(i));
	}

	cout << ans << endl;

	return 0;

}