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 <bits/stdc++.h>
using namespace std;
const int N=200010;
int i,j,k,n,m,x,y,t,b[20];
int fi[N*2],ne[N*2],la[N*2],a[N*2],w[N];
int f[N][20],g[N][20];
void add(int x,int y){
	k++;a[k]=y;
	if (fi[x]==0)fi[x]=k;else ne[la[x]]=k;
	la[x]=k;
}
void dfs(int x,int fa){
	for (j=0;j<20;j++)f[x][j]=1;
	for (int i=fi[x];i;i=ne[i])if (a[i]!=fa){
		dfs(a[i],x);
		for (j=0;j<20;j++)f[x][j]&=f[a[i]][j],g[x][j]+=g[a[i]][j];
	}
	memset(b,0,sizeof b);
	for (int i=fi[x];i;i=ne[i])if (a[i]!=fa)
		for (int j=0;j<20;j++)
			if (g[a[i]][j]&&!f[a[i]][j])
				b[j]++;
	for (j=19;j>=0;j--)if (b[j]>1){
		for (k=j;k<20;k++)if (f[x][k]){w[x]=k;break;}
		break;
	}
	if (!w[x]){for (j=0;j<20;j++)if (f[x][j]){w[x]=max(w[x],j);break;}}
	f[x][w[x]]=0;g[x][w[x]]++;
	for (j=0;j<w[x];j++)f[x][j]=1;
}
int main(){
	scanf("%d",&n);
	for (i=1;i<n;i++){
		scanf("%d%d",&x,&y);
		add(x,y);add(y,x);
	}
	dfs(1,-1);
	int ans=0;
	for (i=1;i<=n;i++)ans=max(ans,w[i]);
//	for (i=1;i<=n;i++)printf("%d ",w[i]);printf("\n");
	printf("%d\n",ans);
	return 0;
}