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;
int read(){
	int x=0,f=1;
	char ch=getchar();
	while (!isdigit(ch)&&ch!='-')
		ch=getchar();
	if (ch=='-')
		f=-1,ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return x*f;
}
int n;
vector <int> e[N];
int Log[N],ans=0,f[N];
void solve(int x,int pre){
	int t=0;
	for (auto y : e[x])
		if (y!=pre){
			solve(y,x);
			t|=f[x]&f[y];
			f[x]|=f[y];
		}
	f[x]^=t;
	int k=t?Log[t]+1:0;
	while (k<30&&(f[x]>>k&1))
		k++;
	ans=max(ans,k);
	f[x]|=1<<k;
	f[x]&=~((1<<k)-1);
}
int main(){
	n=read();
	for (int i=1;i<=n;i++)
		e[i].clear();
	for (int i=1;i<n;i++){
		int a=read(),b=read();
		e[a].push_back(b);
		e[b].push_back(a);
	}
	Log[1]=0;
	for (int i=2;i<=n;i++)
		Log[i]=Log[i>>1]+1;
	solve(1,0);
	cout << ans;
	return 0;
}