Tree is a connected graph without cycles. A leaf of a tree is any vertex connected with exactly one other vertex.

You are given a tree with n vertices and a root in the vertex 1. There is an ant in each leaf of the tree. In one second some ants can simultaneously go to the parent vertex from the vertex they were in. No two ants can be in the same vertex simultaneously except for the root of the tree.

Find the minimal time required for all ants to be in the root of the tree. Note that at start the ants are only in the leaves of the tree.

Input

The first line contains integer n (2 ≤ n ≤ 5·105) — the number of vertices in the tree.

Each of the next n - 1 lines contains two integers xi, yi (1 ≤ xi, yi ≤ n) — the ends of the i-th edge. It is guaranteed that you are given the correct undirected tree.

Output

Print the only integer t — the minimal time required for all ants to be in the root of the tree.

Examples

Input

12
1 2
1 3
1 4
2 5
2 6
3 7
3 8
3 9
8 10
8 11
8 12


Output

6


Input

2
2 1


Output

1
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;

public class E {

	public static void main(String[] args) {

		MyScanner in = new MyScanner();
		int n = in.nextInt();
		List<Integer>[] g = new List[n];
		for(int i=0;i<n;++i){
			g[i] = new ArrayList<>();
		}
		
		for(int i=0;i<n-1;++i){
			int x = in.nextInt()-1;
			int y = in.nextInt()-1;
			g[x].add(y);
			g[y].add(x);
		}
				
		int res = 0;
		for(Integer next:g[0]){
			res = Math.max(res,time(g,next));
		}
		System.out.println(res);
	}

	private static int time(List<Integer>[] g, Integer current) {
		if(g[current].size()==1){
			return 1;
		}
		
		List<Integer> distances = new ArrayList<>();
		Stack<Integer> v = new Stack<Integer>();
		Stack<Integer> d = new Stack<Integer>();
		Set<Integer> visited = new HashSet<>();
		visited.add(0);
		d.push(0);
		v.push(current);
		while(!v.isEmpty()){			
			int curr = v.pop();
			int dst = d.pop();
			
			if(visited.contains(curr)){
				continue;
			}
			
			for(Integer n:g[curr]){
				if(!visited.contains(n)){
					v.push(n);
					d.push(dst+1);
				}
			}
//			System.out.println(curr+":"+dst);
			if(g[curr].size()==1){
				distances.add(dst);
			}
			visited.add(curr);
		}
		
		Collections.sort(distances);
		int t = 1;
		int result = 0;
		for(int i=distances.size()-1;i>=0;--i){
			result = Math.max(result,distances.get(i)+t);
			t++;
		}		
		return result;
	}
	
	  //-----------MyScanner class for faster input----------
	   public static class MyScanner {
	      BufferedReader br;
	      StringTokenizer st;
	 
	      public MyScanner() {
	         br = new BufferedReader(new InputStreamReader(System.in));
	      }
	 
	      String next() {
	          while (st == null || !st.hasMoreElements()) {
	              try {
	                  st = new StringTokenizer(br.readLine());
	              } catch (IOException e) {
	                  e.printStackTrace();
	              }
	          }
	          return st.nextToken();
	      }
	 
	      int nextInt() {
	          return Integer.parseInt(next());
	      }
	 
	      long nextLong() {
	          return Long.parseLong(next());
	      }
	 
	      double nextDouble() {
	          return Double.parseDouble(next());
	      }
	 
	      String nextLine(){
	          String str = "";
		  try {
		     str = br.readLine();
		  } catch (IOException e) {
		     e.printStackTrace();
		  }
		  return str;
	      }

	   }
	   //--------------------------------------------------------

	

}
