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>
#define N 100005
using namespace std;
int mask[N],ans,n;
vector<int> g[N];
void DFS(int u,int fr){
    int two=0,tot=0;
    for(int i:g[u])
        if(i!=fr){
            DFS(i,u);
            two|=tot&mask[i];
            tot|=mask[i];
        }
    int v=31-__builtin_clz(two);
    v= __builtin_ffs(~(tot>>v))-1+v;
    ans=max(ans,v);mask[u]=((tot>>v)<<v)|(1<<v);
}
int main(){
    scanf("%d",&n);
    for(int i=2,f,t;i<=n;i++) 
        scanf("%d%d",&f,&t),g[f].push_back(t),g[t].push_back(f);
    DFS(1,0);
    cout<<ans<<endl;
    return 0;
}