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;
int n,x,y,f[100005][30],ans;
vector <int> a[100005];
void dfs(int x,int fa){
	int z=-1,y;
	for (int i=0;i<a[x].size();++i)
	if (a[x][i]!=fa){
		dfs(y=a[x][i],x);
		for (int j=0;j<30;++j){
			if (f[y][j]&f[x][j]) z=max(z,j);
			f[x][j]|=f[y][j];
		}
	}
	for (++z;z<30;++z)
	if (!f[x][z]) break;
	f[x][z]=1; ans=max(ans,z);
	for (--z;~z;--z) f[x][z]=0;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<n;++i){
		scanf("%d%d",&x,&y);
		a[x].push_back(y);
		a[y].push_back(x);
	}
	dfs(1,0);
	printf("%d\n",ans);
	return 0;
}