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 <functional>
#include <vector>
#define repeat(i, n) for (int i = 0; (i) < int(n); ++(i))
using namespace std;

int main() {
	// input
	int n; scanf("%d", &n);
	vector<vector<int> > g(n);
	repeat (i, n - 1) {
		int a, b; scanf("%d%d", &a, &b); -- a; -- b;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	// solve
	function<int (int, int)> go = [&](int i, int parent) {
		int cnt1 = 0, cnt2 = 0;
		for (int j : g[i]) if (j != parent) {
			int q = go(j, i);
			cnt2 |= cnt1 & q;
			cnt1 |= q;
		}
		int p = 1;
		while (p <= cnt2 or cnt1 & p) p <<= 1;
		return (cnt1 & ~ (p - 1)) | p;
	};
	int p = go(0, -1);
	int k = 31 - __builtin_clz(p);
	// output
	printf("%d\n", k);
	return 0;
}
