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<cstdio>
const int MaxN=200010;
int N;
struct edge{int to;edge*next;}E[MaxN*2],*ne,*fir[MaxN];
void link(int u,int v){*ne=(edge){v,fir[u]};fir[u]=ne++;}
char mx1[MaxN],mn0[MaxN];
int dfs(int i,int f=0){
	int a=0,b=0;
	for(edge*e=fir[i];e;e=e->next)if(e->to!=f){
		int x=dfs(e->to,i);
		b|=a&x;a|=x;
	}
	b=mx1[b];a>>=b;
	return(a>>mn0[a]|1)<<mn0[a]+b;
}
int main(){
	while(scanf("%d",&N)==1){
		ne=E;
		for(int i=1;i<=N;i++)fir[i]=0;
		for(int i=1,u,v;i<N;i++)scanf("%d%d",&u,&v),link(u,v),link(v,u);
		for(int i=1;i<=N;i++){
			mx1[i]=mx1[i/2]+1;
			mn0[i]=i%2?mn0[i/2]+1:0;
		}
		int ans=mx1[dfs(1)]-1;
		printf("%d\n",ans);
	}
}