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=1e5+5;
typedef vector<int>VI;
int n,a,b,i;
vector<int>e[N];
bool f[N][20];
inline void uni(bool*a,bool*b){
	static bool ans[20];memset(ans,0,20);int i=0,x;
	for(;!a[i];)b[i++]=0;
	for(;i<19;++i)
		if(x=a[i]+b[i]+ans[i],x>1)memset(ans,0,i+1),ans[i+1]=1;
			else ans[i]=x;
	memcpy(a,ans,20);
}
void dfs(int u,int fa){
	f[u][0]=1;
	for(int v:e[u])if(v!=fa)
		dfs(v,u),uni(f[u],f[v]);
}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);
	cin>>n;for(i=1;i<n;++i)cin>>a>>b,e[a].push_back(b),e[b].push_back(a);
	dfs(1,0);for(i=19;!f[1][i];--i);cout<<i<<endl;
	return 0;
}