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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
vector<int>pat[100000];
bool flag[100000];
int dfs(int node)
{
	flag[node] = true;
	int d[20];
	fill(d, d + 20, 0);
	for (int i = 0; i < pat[node].size(); i++)
	{
		if (!flag[pat[node][i]])
		{
			int t = dfs(pat[node][i]);
			for (int j = 0; j < 20; j++)if ((t&(1 << j)) != 0)d[j]++;
		}
	}
	int lb = 0;
	for (int i = 0; i < 20; i++)if (d[i] >= 2)lb = i + 1;
	int r = 0;
	for (int i = lb; i < 20; i++)
	{
		if (d[i] == 0)
		{
			r = i;
			break;
		}
	}
	int ret = (1 << r);
	for (int i = r + 1; i < 20; i++)if (d[i]>0)ret += 1 << i;
	return ret;
}
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i < num - 1; i++)
	{
		int za, zb;
		scanf("%d%d", &za, &zb);
		za--, zb--;
		pat[za].push_back(zb);
		pat[zb].push_back(za);
	}
	fill(flag, flag + num, false);
	int r = dfs(0);
	int maxi = 0;
	for (int i = 0; i < 20; i++)if ((r&(1 << i)) != 0)maxi = i;
	printf("%d\n", maxi);
}