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>
#include <cstring>
#include <vector>
#include <ctime>

#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)
#define RepG(i, x) for (int i = head[x]; i; i = edge[i].next)
#define v edge[i].to
#define mp(a, b) make_pair(a, b)

using namespace std;

const int N = 100010;

struct Edge{ int to, next;} edge[N * 2];
int head[N], num;
void add_edge(int a, int b) { edge[++ num] = (Edge){b, head[a]}, head[a] = num;}

int s[N], ans;

void dfs(int x, int f)
{
	int tmp = 0;
	RepG(i, x) if (v != f){
		dfs(v, x);
		int ts = (s[v] & s[x]), t0 = 20;
		if (ts) {
			while (!(ts & (1 << t0))) t0 --; 
			tmp = max(tmp, t0 + 1);
		}
		s[x] |= s[v];
	}
	if (tmp) Rep0(i, tmp - 1) if (s[x] & (1 << i)) s[x] ^= (1 << i);
	for (int i = tmp; i <= 20; i ++){
		if (!(s[x] & (1 << i))){
			s[x] |= (1 << i), ans = max(ans, i);
			break;
		}
		else s[x] ^= (1 << i);
	}
}

int main()
{
	int n;
	scanf("%d", &n);
	Rep(i, n - 1) {
		int a, b;
		scanf("%d%d", &a, &b);
		add_edge(a, b); add_edge(b, a);
	}
	
	dfs(1, 0);
	
	printf("%d\n", ans);
	
	return 0;
}
