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<algorithm>
#define maxn 100005
using namespace std;
int n,ans,f[maxn][30];
struct node { int v; node *nxt; } edge[maxn*2],*head[maxn],*ncnt;
void addedge(int u,int v)
{
	ncnt++;
	ncnt->v=v,ncnt->nxt=head[u];
	head[u]=ncnt;
}
void dfs(int u,int fa)
{
	for(node *p=head[u];p;p=p->nxt)
	{
		int v=p->v;
		if(v==fa) continue;
		dfs(v,u); 
		for(int i=0;i<=20;i++) f[u][i]+=f[v][i];
	}
	int p=0;
	for(int i=20;i>=0;i--)
		if(f[u][i]>1) { p=i+1; break; }
	while(f[u][p]) p++;
	f[u][p]++;
	for(int i=0;i<p;i++) f[u][i]=0;
	ans=max(ans,p);
}
int main()
{
	ncnt=&edge[0];
	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);
	printf("%d\n",ans);
}
