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.

Examples

Input

4 5
1 2
1 3
3 2
2 4
3 4


Output

3


Input

6 3
2 3
4 5
5 6


Output

2


Input

5 8
5 3
2 3
2 4
5 2
5 1
1 4
4 3
1 3


Output

3
import java.util.*;
import java.io.*;

public class Main {
  
  public static void main(String[] abc) {
     Main answer = new Main();
     long solution = answer.solve();
     System.out.println(solution);
  }
  
  private long[] dp;
  
  private long solve() {
     Scanner sc = new Scanner(System.in);     
     int N = sc.nextInt();
     int M = sc.nextInt();
     Map<Integer, List<Integer>> graph = new HashMap<>();
     for(int i=0;i<M;i++) {
        int u = sc.nextInt();
        int v = sc.nextInt();
        List<Integer> children = graph.getOrDefault(u, new ArrayList<>());        
        children.add(v);
        graph.put(u, children);
     }
     long bestLen =0;
     dp = new long[1000000];
     Arrays.fill(dp, -1);
     for(int i=1;i<=N;i++) {
       bestLen = Math.max(bestLen, getBestLen(i, graph));
     }
     return bestLen;
  }
	  
  private long getBestLen(int u, Map<Integer, List<Integer>> graph) {
	 if (dp[u]!=-1) return dp[u];
     if (!graph.containsKey(u)){
        dp[u]=0;
        return dp[u];
     } else {
        long bestChildLen=0;
        for(int child: graph.get(u)){
         	 long currChildLen = getBestLen(child, graph);
             bestChildLen = Math.max(bestChildLen, currChildLen);
        }
		dp[u] = bestChildLen+1;
        return dp[u];
     }
  }
}
