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: Time Limit Exceeded)

#include "bits/stdc++.h"
using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP0(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define REP1(i, n) for (int i = 1, i##_len = (n); i <= i##_len; ++i)

typedef long long LL;
typedef pair<int, int> pii;

const int INTINF = 1e9;
const LL LLINF = 1e18;
#define TENS(n) int(1e##n + n)

// i 番目から出発したときの最長経路
int dp[TENS(5)];
int deg_o[TENS(5)];
int N, M;

void solve()
{
    cin >> N;
    cin >> M;
    // A[i]...iに向かう頂点の配列
    vector<int> A[TENS(5)];
    REP0(i, M)
    {
        int x, y;
        cin >> x >> y;
        A[y].push_back(x);
        deg_o[x]++;
    }

    int ans = 0;
    REP1(i, N)
    {
        if (deg_o[i] == 0) // 終点
        {
            queue<int> S;
            S.push(i);
            while (S.size())
            {
                int v = S.front();
                S.pop();

                for (auto j : A[v])
                { // j : vへ向かう頂点
                    S.push(j);
                    dp[j] = max(dp[j], dp[v] + 1);
                    ans = max(ans, dp[j]);
                }
            }
        }
    }

    cout << ans << endl;
}

int main(int argc, char const *argv[])
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    std::cout << std::fixed << std::setprecision(10);
    solve();
    return 0;
}