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 <string.h>
#include <cstdlib>
#include <vector>
#include <string>
#include <bitset>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
 
typedef long long ll;
typedef long double ld;
 
const ll MOD = 1e9 + 7, INF = 1e18 + 1;
 
using namespace std;

int n;

vector <int> g[1000000];

int label[1000000], f[300000][20], ans;

int dfs (int x, int p)
{
	int mx = 0;
	for (int to : g[x])
		if (to != p)
		{
			dfs (to, x);

			for (int i = 0; i < 20; i++)
			{
				if (f[x][i] && f[to][i]) mx = max (mx, i);
				f[x][i] |= f[to][i];
			}
		}

	while (f[x][mx]) mx++;

	label[x] = mx;

	ans = max (label[x], ans);

	f[x][mx] = 1;

	for (int i = 0; i < mx; i++)
		f[x][i] = 0;

	return label[x];
}

int main ()
{
	cin >> n;

	for (int i = 0; i < n - 1; i++)
	{
		int a, b;

		scanf ("%d%d", &a, &b);

		g[a-1].push_back (b - 1);
		g[b-1].push_back (a - 1);
	}

	dfs (0, -1);

	cout << ans;
}