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
//Wrong answer on test inf.
#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define per(i,x,y) for (int i=(x);i>=(y);i--)
#define ll long long
#define inf 1000000001
#define y1 y1___
using namespace std;
char gc(){
	static char buf[100000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
#define gc getchar
ll read(){
	char ch=gc();ll x=0;int op=1;
	for (;!isdigit(ch);ch=gc()) if (ch=='-') op=-1;
	for (;isdigit(ch);ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
	return x*op;
}
#define N 100005
int n,cnt,head[N],bit[N][25],ans;
struct edge{int to,nxt;}e[N<<1];
void adde(int x,int y){e[++cnt].to=y;e[cnt].nxt=head[x];head[x]=cnt;}
void dfs(int u,int pr){
	for (int i=head[u];i;i=e[i].nxt) if (e[i].to!=pr){
		int v=e[i].to;
		dfs(v,u);
		rep (j,0,20) bit[u][j]+=bit[v][j];
	}
	int Min=0;
	per (i,20,0) if (bit[u][i]>=2){Min=i+1;break;}
	while (bit[u][Min]) Min++;
	ans=max(ans,Min);
	bit[u][Min]=1;
	rep (i,0,Min-1) bit[u][i]=0;
}
int main(){
	n=read();
	rep (i,1,n-1){
		int x=read(),y=read();
		adde(x,y);adde(y,x);
	}
	dfs(1,0);
	printf("%d\n",ans);
	return 0;
}