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;
#define N 100010
int he[N],ne[N<<1],e[N<<1],cnt,bit[N],mark[N],ans,n;
void dfs(int x,int fa){
    for (int i=he[x];i;i=ne[i]){
        if (e[i]!=fa) dfs(e[i],x);
    }
    int S1=0,S2=0,S3=0;
    for (int i=he[x];i;i=ne[i]){
        S3=S1&bit[e[i]];
        S2|=S3;
        S1|=bit[e[i]];
    }
    S1^=S2;
    for (int i=0;i<20;i++){
        if ((1<<i)>S2 && !(S1&(1<<i))){
            mark[x]=i;
            ans=max(ans,i);
            break;
        }
    }
    bit[x]=(S1>>mark[x])<<mark[x];
    bit[x]|=(1<<mark[x]);
}
void add(int u,int v){ne[++cnt]=he[u];he[u]=cnt;e[cnt]=v;}
int main(){
    scanf("%d",&n);
    for (int i=1;i<n;i++){
        int u,v; scanf("%d%d",&u,&v);
        add(u,v); add(v,u);
    }
    dfs(1,0);
    printf("%d\n",ans);
    return 0;
}