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 MAXN = 100 * 1000 + 23, MXLN = 17 + 3;











int n, mx, dp[MAXN], cnt[MAXN][MXLN];
vector<int> g[MAXN];


void dfs(int v = 0, int p = -1) {
	for (auto u : g[v]) if (u != p) {
		dfs(u, v);
		dp[v] |= dp[u];
		for (int i = 0; i < MXLN; i++) cnt[v][i] += ((dp[u] >> i) & 1);
	}
//	/*
	int b = 0, f = 1;
	for (int i = 0; i < MXLN; i++) {
		if (cnt[v][i] > 1) {
			b = i + 1;
			f = 1;
		} else {
			if (cnt[v][i]) b += f;
			else f = 0;
		}
	}
//	*/
//	for (int i = 0; i < MXLN; i++) if (cnt[v][i] > 1) b = i;
	dp[v] &= (((1 << MXLN) - 1) ^ ((1 << b) - 1));
	dp[v] |= (1 << b);
	mx = max(mx, b);
}




int main() {
	cin >> n;
	for (int i = 1; i < n; i++) {
		int u, v;
		cin >> u >> v;
		g[--u].push_back(--v), g[v].push_back(u);
	}
	dfs();
	cout << mx;
	return 0;
}