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 <cstdlib>
#include <cstring>
#include <vector>
#define MAXN 100010
int n, ans, f[MAXN];
std::vector<int> next[MAXN];
void gao(int x, int fa) {
	int tmp = 0, all = 0;
	for (int y : next[x])
		if (y != fa) gao(y, x), tmp |= all & f[y], all |= f[y];
	if (!all) {
		f[x] = 1;
		ans = std::max(ans, 1);
	} else {
		int count = 0;
		while (((1 << count) < tmp) || ((all >> count) & 1)) count++;
		f[x] = (all >> count << count) | (1 << count);
		ans = std::max(ans, count);
	}
}
int main() {
	scanf("%d", &n);
	for (int u, v, i = 1; i < n; i++) scanf("%d%d", &u, &v), next[u].push_back(v), next[v].push_back(u);
	gao(1, 0);
	printf("%d\n", ans);
	return 0;
}