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;
typedef pair<int,int> PII;
const int N = 100010;
int n,x,y;
vector<int> v[N],f[N];

void dfs(int x,int fa){
	int t;
	vector<int> tmp;
	for(int i=0;i<v[x].size();i++) if((t=v[x][i])!=fa){
		dfs(t,x);
		for(int j=0;j<f[t].size();j++){
			if(j>=tmp.size()) tmp.push_back(f[t][j]); 
			else tmp[j] += f[t][j];
		}
	}
	int l = tmp.size();
	for(int i = tmp.size()-1;i>=0;i--){
		if(tmp[i]==0) l = i;
		if(tmp[i]>1) {
			break;
		}
	}
	if(l>=tmp.size()) tmp.push_back(1); else tmp[l]=1;
	for(int i=l-1;i>=0;i--) tmp[i] =0;
	f[x] = tmp;
}

int main(){
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
		scanf("%d%d",&x,&y);
		v[x].push_back(y);
		v[y].push_back(x);
	}
	dfs(1,0);
	printf("%d\n",f[1].size()-1);
	return 0;
}