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 n;
const int N = 200000;
vector <int> bi[N];
int dfs(int t, int f = 0)
{
    int c = 0, d = 0;
    for (int i = 0; i < bi[t].size(); ++ i)
        if (bi[t][i] != f)
        {
            int v = dfs(bi[t][i], t);
            d |= c & v;
            c ^= v;
        }
    for (int i = 30; ~i; -- i)
        if (i == 0 || (d & (1 << (i - 1))))
            for (int j = i; ; ++ j)
                if (!(c & (1 << j)))
                    return (c | (1 << j)) ^ (c & ((1 << j) - 1));
}
int main()
{
    cin >> n;
    for (int i = 1; i < n; ++ i)
    {
        int a, b;
        cin >> a >> b;
        bi[a].push_back(b);
        bi[b].push_back(a);
    }
    int v = dfs(1);
    for (int i = 30; ~i; -- i)
        if (v & (1 << i))
        {
            cout << i << endl;
            return 0;
        }
}
