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 + 5;
int n;
int a , b;
vector < int > v[N];
int dfs(int node , int parent){
    int mask1 = 0;
    int mask2 = 0;
    for(int next : v[node]){
        if(next != parent){
            int val = dfs(next , node);
            mask2 |= val & mask1;
            mask1 |= val;
        }
    }
    int mn = 0;
    for(int i = 0 ; i < 20 ; ++i){
        if((mask2 >> i) & 1){
            mn = i;
        }
    }
    for(int i = mn ; i < 20 ; ++i){
        if(!((mask1 >> i) & 1)){
            mask1 |= 1 << i;
            mask1 &= ~((1 << i) - 1);
            break;
        }
    }
    return mask1;
}
int main(){
    scanf("%d" , &n);
    for(int i = 1 ; i < n ; ++i){
        scanf("%d %d" , &a , &b);
        v[a].emplace_back(b);
        v[b].emplace_back(a);
    }
    int ans = dfs(1 , 0);
    for(int i = 19 ; ; --i){
        if((ans >> i) & 1){
            printf("%d\n" , i);
            break;
        }
    }
}