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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
 
#define For(i, j, k) for(int i = j; i <= k; i++)
#define Forr(i, j, k) for(int i = j; i >= k; i--)
#define pb push_back
 
using namespace std;
 
const int N = 100010;
 
vector<int> G[N];
int n, mask[N];
 
void DFS(int o, int f){
	int label = 0, s = 0;
	//for(int v : G[o]){
	for (int i=0;i<G[o].size();++i)
	{
	int v=G[o][i];
		if(v == f) continue;
		DFS(v, o);
		label |= s & mask[v];// 值域> 
		s |= mask[v];//子树的前缀值域 或 
	}
	label = label << 1 | 1;
	Forr(i, 30, 0) if(label & (1 << i))
		For(j, i, 30) if(!(s & (1 << j))){//不能等于s中的值 
			mask[o] = //(s ^ (s & ((1 << j) - 1))) | (1 << j);
			  (s>>j<<j)|(1<<j);
			return;
		}
}
 
int main(){
	scanf("%d", &n);
	For(i, 2, n){
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].pb(v), G[v].pb(u);
	}
	DFS(1, 0);
	//for(int i=1;i<=n;++i) printf("%d\n",mask[i]) ;
	Forr(i, 30, 0) if(mask[1] & (1 << i)){
		printf("%d\n", i);
		return 0;
	}
}