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<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define fo(i,a,b) for(i=a;i<=b;i++)
#define min(x,y) ((x)<(y)?(x):(y))
#define max(x,y) ((x)>(y)?(x):(y))

using namespace std;

const int maxn=2e5+5;

int fi[maxn],ne[maxn*2],dui[maxn*2],qc[maxn],f[maxn*10];
int cc[maxn];
int i,j,k,l,m,n,x,y,now,ans;

void add(int x,int y){
	if (fi[x]==0) fi[x]=++now; else ne[qc[x]]=++now;
	dui[now]=y; qc[x]=now;
}
void dfs(int x,int y){
	int i,t=0;
	for(i=fi[x];i;i=ne[i]){
		if (dui[i]==y) continue;
		dfs(dui[i],x);
		t=max(t,f[x] & f[dui[i]]);
		f[x]=f[x] | f[dui[i]];
	}
	fo(i,1,24) if (cc[i]>t && ((f[x] & cc[i])==0)){
		t=i; break;
	}
	ans=max(ans,t);
	f[x]=(f[x]&(cc[24]-cc[t]))|cc[t];
}
int main(){
//	freopen("009d.in","r",stdin);
	scanf("%d",&n);
	fo(i,1,n-1){
		scanf("%d%d",&x,&y);
		add(x,y);add(y,x);
	}
	cc[1]=1;
	fo(i,2,24) cc[i]=cc[i-1]*2;
	dfs(1,0);
	printf("%d\n",ans-1);
	return 0;
}