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;

vector<int> g[N];
int ant[N];
int n;


static int compute(int arg) {
	for (int lg = 0; ; ++lg)
		if ((1 << lg + 1) > arg)
			return lg;  }

static int dfs(int u, int far = 0) {
	int viz(0), force(0);

	for (auto v: g[u]) if (v != far) {
		int msk = dfs(v, u);
		force|= (viz & msk);
		viz|= msk; }

	ant[u] = compute(force);
	while ((1 << ant[u]) & viz)
		++ant[u];

	viz = viz & ~((1 << ant[u]) - 1);
	viz|= 1 << ant[u];

	return viz; }

int main() {
#ifdef HOME
	freopen("agc9d.in", "r", stdin);
	freopen("agc9d.out", "w", stdout);
#endif
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	cin >> n;
	for (int u, v, i = 1; i < n; ++i) {
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u); }

	dfs(1);
	cout << *max_element(ant + 1, ant + n + 1) << endl;

	return 0; }

