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>
const int N=1e5+7;
struct edge{
	int to,nx;
}e[N*2];
int n,e0[N],ep=2,f[N];
void f1(int w,int pa){
	for(int i=e0[w];i;i=e[i].nx){
		int u=e[i].to;
		if(u!=pa)f1(u,w);
	}
	int v1=0,v2=0,v3=1;
	for(int i=e0[w];i;i=e[i].nx){
		int u=e[i].to;
		if(u!=pa){
			int x=f[u];
			v2|=v1&x;
			v1|=x;
		}
	}
	for(;v3<=v2||(v3&v1);v3<<=1);
	f[w]=(v1|v3)&~(v3-1);
}
int main(){
	scanf("%d",&n);
	for(int i=1,a,b;i<n;++i){
		scanf("%d%d",&a,&b);
		e[ep]=(edge){b,e0[a]};e0[a]=ep++;
		e[ep]=(edge){a,e0[b]};e0[b]=ep++;
	}
	f1(1,0);
	int ans=0;
	for(;f[1]>1;f[1]>>=1,++ans);
	printf("%d\n",ans);
	return 0;
}