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>
#define N 100009
using namespace std;

int n,ans,tot,fst[N],pnt[N<<1],nxt[N<<1],fa[N],tg[N];
void add(int x,int y){
	pnt[++tot]=y; nxt[tot]=fst[x]; fst[x]=tot;
}
void dfs(int x){
	int i,j,tmp,y;
	for (i=fst[x]; i; i=nxt[i]){
		y=pnt[i];
		if (y!=fa[x]){
			fa[y]=x; dfs(y);
			tmp=tg[x]&tg[y];
			for (j=1; j<=tmp; j<<=1) tg[x]|=j;
			tg[x]|=tg[y];
		}
	}
	for (i=1; tg[x]&i; i<<=1) tg[x]^=i;
	tg[x]|=i;
	ans=max(ans,tg[x]);
}
int main(){
	scanf("%d",&n);
	int i,x,y;
	for (i=1; i<n; i++){
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	dfs(1);
	for (i=-1; ans; ans>>=1) i++;
	printf("%d\n",i);
	return 0;
}
