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<algorithm>
using namespace std;
int n,u,v,ans,dp[110000][33];
int edgenum,vet[210000],Next[210000],Head[110000];
void addedge(int u,int v){
	vet[++edgenum]=v;
	Next[edgenum]=Head[u];
	Head[u]=edgenum;
}
void dfs(int u,int fa){
	int v;
	for (int e=Head[u];e;e=Next[e]){
		v=vet[e];
		if (v!=fa){
			dfs(v,u);
			for (int i=0;i<=30;i++) dp[u][i]+=dp[v][i];
		}
	}
	v=0;
	for (int i=0;i<=30;i++)
		if (dp[u][i]>1) v=i+1;
	while (dp[u][v]) v++;
	ans=max(ans,v); dp[u][v]=1;
	for (int i=0;i<v;i++) dp[u][i]=0;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<n;i++){
		scanf("%d%d",&u,&v);
		addedge(u,v); addedge(v,u);
	}
	dfs(1,0);
	printf("%d\n",ans);
	return 0;
}
