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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;

inline int getint()
{
	static char c;
	while ((c = getchar()) < '0' || c > '9');

	int res = c - '0';
	while ((c = getchar()) >= '0' && c <= '9')
		res = res * 10 + c - '0';
	return res;
}

const int MaxN = 100000;

int n;
vector<int> adj[MaxN + 1];

int sta[MaxN + 1];
int lab[MaxN + 1];

int dfs(int u, int prev)
{
	sta[u] = 0;

	int mask = 0;
	for (int v : adj[u])
		if (v != prev)
		{
			int sv = dfs(v, u);
			mask |= sv & sta[u];
			sta[u] |= sv;
		}

	if (sta[u] == 0)
		sta[u] = 1, lab[u] = 0;
	else
	{
		int l = 0;
		while (1 << l <= mask)
			++l;
		while (sta[u] >> l & 1)
			++l;

		lab[u] = l;
		sta[u] >>= l, sta[u] <<= l;
		sta[u] |= 1 << l;
	}

	return sta[u];
}

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

	dfs(1, 0);

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

	return 0;
}