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<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

int main(){
    int n, m;
    cin >> n >> m;

    vector<int> child[n], dp(n, -1);
    int x, y;

    for(int i = 0; i < m; i++){
        cin >> x >> y;
        child[--x].push_back(--y);
    }

    queue<pair<int,int>> q;
    for(int i = 0; i < n; i++)  q.push({i,0});

    while(!q.empty()){
        pair<int,int> p = q.front();    q.pop();
        if(dp[p.first] >= p.second) continue;
        dp[p.first] = p.second;
        for(int c : child[p.first]){
            if(dp[p.first]+1 > dp[c])   q.push({c, dp[p.first]+1});
        }
    }

    cout << *max_element(dp.begin(), dp.end()) << endl;

    return 0;
}