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
//In the Name of God
//Let Our Voices Echo
#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define X real()
#define Y imag()
typedef long long ll;
typedef double ld; 
typedef complex<ld> point;
const ld eps=1e-9;
const int MAX=1e5+9,MOD=1e9+7,lg=20;
int n,ans;
vector<int> g[MAX];
int dfs(int v,int p=-1)
{
	int res=0,mi=0;
	for (auto u:g[v])	
		if (u!=p)
		{
			int mask=dfs(u,v);
			for (int i=0;i<lg;i++) 
				if ((mask>>i&1) && (res>>i&1)) 
					mi=max(mi,i+1);
			res|=mask;
		}
	for (int i=mi;;i++)
		if (!(res>>i&1))
		{
			ans=max(ans,i);
			res|=(1<<i);
			for (int j=i-1;j>=0;j--)
				if (res>>j&1)
					res^=(1<<j);
			return res;
		}
	
}
int main()
{
	ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cin>>n;
	for (int i=1,v,u;i<n;i++) cin>>v>>u,v--,u--,g[v].push_back(u),g[u].push_back(v);
	dfs(0);
	cout<<ans;
	return 0;
}