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 <cstdio>
#include <vector>
#define pb push_back
std::vector<int> adj[100005];
int app[100005][25];
int dfs(int u, int fa = -1)
{
	int ans = 0;
	for (int v : adj[u])
	{
		if (v == fa)
			continue;
		ans = std::max(ans, dfs(v, u));
		for (int j = 0; j < 20; j++)
			app[u][j] += app[v][j];
	}
	int res = 0;
	for (int i = 19; i >= 0; i--)
	{
		if (app[u][i] > 1)
		{
			res = i + 1;
			break;
		}
	}
	while (app[u][res])
		res++;
	for (int j = 0; j < res; j++)
		app[u][j] = 0;
	app[u][res] = 1;
	return std::max(ans, res);
}
int main()
{
	// freopen("AGC009-D.in", "r", stdin);
	int n;
	scanf("%d", &n);
	for (int i = 1; i < n; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		adj[--u].pb(--v);
		adj[v].pb(u);
	}
	printf("%d\n", dfs(0));
	return 0;
}