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 EXP = 30;
const int DIM = 200005;

vector<int> edg[DIM];

int dfs(int x, int f, int &ans)
{
    vector<int> aux(EXP, 0);
    
    for (int y : edg[x]) if (y != f) {
        int val = dfs(y, x, ans);
       
        for (int i = 0; i < EXP; ++i)
            aux[i] += ((val >> i) & 1);
    }
    
    int pos = -1;
    for (int i = 0; i < EXP; ++i)
        if (aux[i] >= 2) pos = i;
    
    for (int i = pos + 1; i < EXP; ++i) if (!aux[i]) {
        int msk = (1 << i);
        for (int j = i + 1; j < EXP; ++j)
            msk ^= (1 << j) * aux[j];
        
        ans = max(ans, i);
        return msk;
    }
    
    return 0;
}


int main(void)
{
    int n; cin >> n;
    for (int i = 1; i < n; ++i) {
        int x, y;
        cin >> x >> y;
        
        edg[x].push_back(y);
        edg[y].push_back(x);
    }
 
    int ans = 0; dfs(1, 0, ans);
    cout << ans << endl;
    
    return 0;
}