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>
#define ll long long
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
int f[100100][20],n,ans;
vector<int> e[101000];
void dfs(int x,int dad){
	int lim[20],now=0;memset(lim,0,sizeof(lim));
	for(int v:e[x]) if(v!=dad){
		dfs(v,x);
		for(int j=0;j<20;j++){
			if(lim[j]&&f[v][j]) now=max(now,j+1);
			lim[j]|=f[v][j];
		}
	}
	while(lim[now]) now++;
	ans=max(ans,now);
	for(int i=0;i<now;i++) lim[i]=0;
	lim[now]=1;
	memcpy(f[x],lim,sizeof(lim));
}
int main(){
	read(n);
	for(int i=1;i<n;i++){
		int x,y;read(x);read(y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs(1,0);
	cout<<ans<<'\n';
	return 0;
}
