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=1e5+10;
int gi() {
    int x=0,o=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
struct edge { int to,nxt; } e[N<<1];
int tot=1,h[N];
void adde(int u,int v) {
    e[++tot]=(edge){v,h[u]};h[u]=tot;
    e[++tot]=(edge){u,h[v]};h[v]=tot;
}
int n,ans=0,f[N][30];
void dfs(int u,int fa) {
    for(int i=h[u];i;i=e[i].nxt) {
        int v=e[i].to; if(v==fa) continue;
        dfs(v,u);
        for(int j=20;j>=0;j--) f[u][j]+=f[v][j];
    }
    int p=0;
    for(int i=20;i>=0;i--)
        if(f[u][i]>1) { p=i+1;break; }
    while(f[u][p]) p++;
    f[u][p]++;
    for(int j=0;j<p;j++) f[u][j]=0;
    ans=max(ans,p);
}
int main() {
    cin>>n;
    for(int i=1,u,v;i<n;i++) u=gi(),v=gi(),adde(u,v);
    dfs(1,0);printf("%d\n",ans);
    return 0;
}
//orzgzy
//鸡贼明年进队超稳
