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=100005;
struct edge{
	int to,next;
}e[N*2];
int head[N],tot;
int f[N],n,ans;
void add(int x,int y){
	e[++tot]=(edge){y,head[x]};
	head[x]=tot;
}
void dfs(int x,int fa){
	int has=0,bad=0;
	for (int i=head[x];i;i=e[i].next)
		if (e[i].to!=fa){
			dfs(e[i].to,x);
			bad|=(has&f[e[i].to]);
			has|=f[e[i].to];
		}
	int tmp=16;
	for (;tmp>=0;tmp--)
		if (bad&(1<<tmp)) break;
	for (tmp++;has&(1<<tmp);tmp++);
	f[x]=(has&((1<<17)-(1<<tmp)))|(1<<tmp);
	ans=max(ans,tmp);
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<n;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	dfs(1,0);
	printf("%d\n",ans);
}