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<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#define SF scanf
#define PF printf
#define MAXN 100010
using namespace std;
int n,u,v,ans;
vector<int> a[MAXN];
int bit[MAXN],f[MAXN];
void dfs(int x,int fa){
	int l=0;
	for(int i=0;i<int(a[x].size());i++){
		int u=a[x][i];
		if(u==fa)
			continue;
		dfs(u,x);
		l|=(bit[x]&bit[u]);
		bit[x]|=bit[u];
	}
	if(bit[x]==0)
		bit[x]=1;
	else{
		while((1<<f[x])<l||(1<<f[x])&bit[x])
			f[x]++;
		bit[x]=(bit[x]/(1<<f[x])*(1<<f[x]))|(1<<f[x]);
	}
	ans=max(ans,f[x]);
}
int main(){
	SF("%d",&n);
	for(int i=1;i<n;i++){
		SF("%d%d",&u,&v);	
		a[u].push_back(v);
		a[v].push_back(u);
	}
	dfs(1,0);
	PF("%d",ans);
}	