We will recursively define uninity of a tree, as follows: (Uni is a Japanese word for sea urchins.)

* A tree consisting of one vertex is a tree of uninity 0.
* Suppose there are zero or more trees of uninity k, and a vertex v. If a vertex is selected from each tree of uninity k and connected to v with an edge, the resulting tree is a tree of uninity k+1.



It can be shown that a tree of uninity k is also a tree of uninity k+1,k+2,..., and so forth.

You are given a tree consisting of N vertices. The vertices of the tree are numbered 1 through N, and the i-th of the N-1 edges connects vertices a_i and b_i.

Find the minimum k such that the given tree is a tree of uninity k.

Constraints

* 2 ≦ N ≦ 10^5
* 1 ≦ a_i, b_i ≦ N(1 ≦ i ≦ N-1)
* The given graph is a tree.

Input

The input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}


Output

Print the minimum k such that the given tree is a tree of uninity k.

Examples

Input

7
1 2
2 3
2 4
4 6
6 7
7 5


Output

2


Input

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


Output

3
#include "bits/stdc++.h"
using namespace std;
const int N = 1e5 + 5;
int n;
int a , b;
vector < int > v[N];
int dfs(int node , int parent){
	int curor = 0;
	int two = 0;
	for(int next : v[node]){
		if(next == parent){
			continue;
		}
		int tmp = dfs(next , node);
		two |= tmp & curor;
		curor |= tmp;
	}
	two |= two >> 8;
	two |= two >> 4;
	two |= two >> 2;
	two |= two >> 1;
	int allowed = (~two) & (~curor);
	int me = allowed & -allowed;
	return (curor | me) & (~(me - 1));
}
int main(){
	scanf("%d" , &n);
	for(int i = 1 ; i < n ; ++i){
		scanf("%d %d" , &a , &b);
		v[a].emplace_back(b);
		v[b].emplace_back(a);
	}
	int ans = dfs(1 , 0);
	for(int i = 16 ; ; --i){
		if((ans >> i) & 1){
			printf("%d\n" , i);
			break;
		}
	}
}