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>
#include <algorithm>

using namespace std;

struct Edge {
  int t,next;
  Edge() {}
  Edge(int a,int b):t(a),next(b) {}
}; 

Edge e[200005];
int head[100005];

int f[100005],g[100005],ans;

void dfs(int x,int fa) {
  int sum=0;
  for(int i=head[x];i;i=e[i].next)
    if (e[i].t!=fa) {
    	int u=e[i].t;
    	dfs(u,x);
    	sum|=(g[x]&g[u]);
    	g[x]|=g[u];
	}
  for(int i=0;i<30;i++)
    if (!((1<<i)<sum)&&!(g[x]&(1<<i))) {
    	f[x]=i;
    	break;
	} 
  g[x]=((g[x]>>f[x]<<f[x])|(1<<f[x]));
  ans=max(ans,f[x]);
}

int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<n;i++) {
  	int x,y;
  	scanf("%d%d",&x,&y);
  	e[2*i-1]=Edge(y,head[x]);
  	head[x]=2*i-1;
  	e[2*i]=Edge(x,head[y]);
  	head[y]=2*i;
  }
  dfs(1,0);
  printf("%d\n",ans);
  return 0;
}