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 <cstdio>
#include <algorithm>
int head[100001], next[199999], to[199999], q[100001], fa[100001], N, E, f[100001][20];
int main()
{
	scanf("%d", &N);
	for (int i = 1, u, v; i < N; i++)
	{
		scanf("%d%d", &u, &v);
		next[++E] = head[u], to[E] = v, head[u] = E;
		next[++E] = head[v], to[E] = u, head[v] = E;
	}
	int H = 0, T = 1, u;
	q[1] = 1;
	while (H < T)
		for (int e = head[u = q[++H]]; e; e = next[e])
			if (to[e] != fa[u])
				fa[q[++T] = to[e]] = u;
	for (int i = N; i; i--)
	{
		int u = q[i];
		for (int e = head[u]; e; e = next[e])
			if (to[e] != fa[u])
				for (int j = 0; j < 20; j++)
					f[u][j] += f[to[e]][j];
		f[u][0]++;
		int z = -1;
		for (int j = 0; j < 19; j++)
			if (f[u][j] > 1)
			{
				z = j;
				f[u][j] = 0;
				f[u][j + 1]++;
			}
		for (int j = 0; j <= z; j++)
			f[u][j] = 0;
	}
	int D = 19;
	while (!f[1][D])
		D--;
	printf("%d\n", D);
	return 0;
}
