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;

int nth_bit(int64_t num, int n){
    return (num >> n) & 1;
}

int N;
vector<int> edges[100000];

int dfs(int i, int p){
    vector<int> values;
    for(int j : edges[i]) if(j != p) values.push_back(dfs(j, i));

    int myvalue = -1;
    for(int k=0; k<30; k++){
        int num = 0;
        for(int n : values) if(nth_bit(n, k)) num++;
        if(num >= 2){
            myvalue = -1;
        }else if(num == 0 && myvalue == -1){
            myvalue = 1<<k;
        }
    }
    int ret = myvalue;
    for(int n : values) ret |= n;
    ret = ret / myvalue * myvalue;
    return ret;
}

int main(){
    cin >> N;
    for(int i=0; i<N-1; i++){
        int a, b;
        cin >> a >> b;
        edges[a-1].push_back(b-1);
        edges[b-1].push_back(a-1);
    }

    int ret = dfs(0, -1);
    for(int k=29; k>=0; k--) if(nth_bit(ret, k)){
        cout << k << endl;
        return 0;
    }
}