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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    int lp[n+1]; //lp[i]はiで終わるlongest path
    int start[n+1]; //start[i]はlp[i]のstart node
    //initialize
    for (int i = 0; i < n+1; i++)
    {
        lp[i]=0;
        start[i]=0;
    }
    for (int count = 0; count < m; count++)
    {
        int x,y;
        scanf("%d %d",&x,&y);
        //x->yをlp[x]の後ろにつけた場合
        if(lp[y]<lp[x]+1)
        {
            lp[y]=lp[x]+1;
            if(start[x]==0)start[y]=x;
            else start[y]=start[x];
        }
        //x->yをstart[i]=yなるlp[i]の前につけた場合
        for (int i = 1; i < n+1; i++)
        {
            if(start[i]==y)
            {
                lp[i]+=1;
                start[i]=x;
            }
        }
    }
    int max=0;
    for (int i = 1; i < n+1; i++)
    {
        if(lp[i]>max)max=lp[i];
    }
    printf("%d\n",max);
    return 0;
}