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;

vector<int> child[100005];
vector<int> adjl[100005];

int ans = 0;

int func(int node){
if (child[node].size()==0) return 1;
vector<int> v;
for (int x : child[node]){
    v.push_back(func(x));
}
int mx = -1;
for (int x = 0; x<23; x++){
    bool c = false;
    for (int y : v){
        if ((y&(1<<x))!=0){
            if (c) mx = x;
            else c = true;
        }
    }
}
for (int x = mx+1; x<23; x++){
    bool can = true;
    int t = 0;
    for (int y : v){
        if ((y&(1<<x))!=0){
            can = false;
            break;
        }
        t = (t|y);
    }
    if (can){
        ans = max(ans,x);
        return (t&(~((1<<(x))-1)))|(1<<x);
    }
}
}

void dfs(int node, int parent){
    for (int x : adjl[node]){
        if (x==parent) continue;
        child[node].push_back(x);
        dfs(x,node);
    }
}

int main(){
int n;
scanf("%d",&n);
for (int x = 0; x<n-1; x++){
    int a,b;
    scanf("%d%d",&a,&b);
    adjl[a].push_back(b);
    adjl[b].push_back(a);
}
dfs(1,-1);
func(1);
printf("%d",ans);
}
