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=2e5+10;
int n,w[N],head[N],nxt[N];
void add(int f,int t){
	static int cnt=0;
	w[++cnt]=t;
	nxt[cnt]=head[f];
	head[f]=cnt;
}
//问题等价于给点标号,若k(u)=k(v),则(u,v)路径上存在一点x使得k(x)>k(v) 
//贪心选择最小的标号 
int S[N],ans;
void dfs(int x,int fa){
	int K=0;
	for (int i=head[x];i;i=nxt[i]){
		int v=w[i];
		if (v==fa) continue;
		dfs(v,x);
		for (int i=0;i<20;i++)
		if ((S[x]&S[v])>>i&1) K=max(K,i+1);
		S[x]|=S[v];
	}
	while (S[x]>>K&1) K++;
	ans=max(ans,K);
	S[x]&=~((1<<K)-1);
	S[x]|=(1<<K);
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);add(v,u);
	}
	dfs(1,0);
	printf("%d\n",ans);
	return 0;
}