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=200005;
int n,x,y,tot,ne[N],fi[N],zz[N];
void jb(int x,int y){
	ne[++tot]=fi[x];
	fi[x]=tot;
	zz[tot]=y;
}
int dfs(int x,int y){
	int s1=0,s2=0,p=0;
	for (int k=fi[x];k;k=ne[k])
		if (zz[k]!=y){
			int p=dfs(zz[k],x);
			s2|=s1&p;
			s1|=p;
		}
	while (((1<<p)<s2)||(s1&(1<<p)))p++;
	return (s1>>p<<p)|(1<<p);
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<n;i++){
		scanf("%d%d",&x,&y);
		jb(x,y);jb(y,x);
	}
	printf("%d\n",(int)log2(dfs(1,0)));
}