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 <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=100010;
int tt,ans;
int f[N],head[N],to[N*2],nxt[N*2];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
inline void dfs(int x,int fa) {
    int s=0,t=0;
    for(int i=head[x];i;i=nxt[i]) {
	int y=to[i];
	if(y!=fa) {
	    dfs(y,x);
	    s|=f[x]&f[y],f[x]|=f[y];
	}
    }
    while((1<<t)<=s||f[x]>>t&1) ++t;
    ans=max(ans,t),f[x]=((f[x]>>t)|1)<<t;
}
int main() {
    int n;
    cin>>n;
    for(int i=1;i<n;i++) {
	int x=gi(),y=gi();
	to[++tt]=y,nxt[tt]=head[x],head[x]=tt;
	to[++tt]=x,nxt[tt]=head[y],head[y]=tt;
    }
    dfs(1,0);
    cout<<ans;
    return 0;
}