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;
#define mid ((s + e) / 2)
#define makefast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define comp(x) (lower_bound(vec.begin(), vec.end(), x) - vec.begin())
#define lc (v * 2)
#define rc (v * 2 + 1)
long long M = 1e9 + 7;
//>>>>>>>>>>>>>>>>>>>

const int N = 1e5 + 7;

vector <int> ad[N];
int n;
int cnt[N][30];
int res = 0;
void solve(int v, int p = 0) {
	if (ad[v].size() == 1 && ad[v].back() == p) {
		cnt[v][0] = 1;
		return;
	}
	for (auto u: ad[v])
		if (u != p) {
			solve(u, v);
			for (int i = 0; i < 30; i++)
			   cnt[v][i] += cnt[u][i];	
		}
	int lst = 29;
	for (int i = 28; i >= 0; i--)
		if (cnt[v][i] > 1)
			break;
		else if (!cnt[v][i])
			lst = i;
	for (int i = 0; i < lst; i++)
		cnt[v][i] = 0;
	cnt[v][lst] = 1;
	res = max(res, lst);
}
int main() {
	cin >> n;
	for (int i = 1; i < n; i++) {
		int v, u;
		cin >> v >> u;
		ad[v].push_back(u);
		ad[u].push_back(v);
	}
	solve(1);
	cout << res;

	return 0;
}

