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>
#include <cstring>
#define MAXN 1000010
#define MAXD 21

struct edge{
	int to,next;
	edge(int _to=0,int _next=0):to(_to),next(_next){}
}e[MAXN<<1];

int n;
int g[MAXN],nume;
int f[MAXN][MAXD];

void addEdge(int u,int v){
	e[nume]=edge(v,g[u]);
	g[u]=nume++;
}

void dfs(int x,int p){
	for(int i=g[x];~i;i=e[i].next)
		if(e[i].to^p)
			dfs(e[i].to,x);
	static int cnt[MAXD];
	memset(cnt,0,sizeof cnt);
	for(int i=g[x];~i;i=e[i].next)
		if(e[i].to^p)
			for(int j=0;j<MAXD;j++)
				if(f[e[i].to][j])
					cnt[j]++;
	int pos=MAXD-1;
	while(pos>=0 && cnt[pos]<=1) pos--;
	while(cnt[++pos]);
	f[x][pos]=1;
	for(int i=pos+1;i<MAXD;i++)
		f[x][i]=cnt[i];
}

int main(){
#ifdef DEBUG
	freopen("D.in","r",stdin);
#endif
	memset(g,-1,sizeof g);
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		addEdge(u,v);
		addEdge(v,u);
	}
	dfs(1,0);
	for(int i=MAXD-1;i>=0;i--)
		if(f[1][i]){
			printf("%d\n",i);
			return 0;
		}
	return 0;
}
