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
/*
神题
从叶结点向根贪心，每个点维护二进制集合，贪心合并子树
答案O(logn)-->O(n*logn)
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 100010
using namespace std;

struct edge{int x, y, next;}a[N*2];
int n, bin[40], mx, f[N], sum[40], l, p[N], x, y;

inline void add(int x, int y){a[++l].x=x; a[l].y=y; a[l].next=p[x]; p[x]=l;}

inline void dfs(int x, int fa){
	for(int i=p[x]; i; i=a[i].next)if(fa!=a[i].y)dfs(a[i].y, x);
	memset(sum, 0, sizeof(sum));
	for(int i=p[x]; i; i=a[i].next)if(fa!=a[i].y){
		for(int j=0; j<=30; j++)if(f[a[i].y]&bin[j])sum[j]++;
	}
	int now=31; f[x]=0;
	for(int i=30; i>=0; i--){
		if(!sum[i]){for(int j=i+1; j<=now-1; j++)f[x]+=bin[j]; now=i;}
		if(sum[i]>=2)break;
	}
	f[x]+=bin[now];
}

int main(){
	scanf("%d", &n);
	bin[0]=1; for(int i=1; i<=30; i++)bin[i]=bin[i-1]<<1;
	l=0; memset(p, 0, sizeof(p));
	for(int i=1; i<=n-1; i++){scanf("%d%d", &x, &y); add(x, y); add(y, x);}
	dfs(1, 0);
	memset(sum, 0, sizeof(sum));
	for(int i=0; i<=30; i++)if(f[1]&bin[i])sum[i]++;
	mx=-1; for(int i=0; i<=30; i++)if(sum[i])mx=i;
	printf("%d", mx);
	return 0;
}