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 <vector>
using namespace std;

#define NR 101000

vector<int> g[NR];
int n;

int dfs(int x, int fa) {
	int mask = 0;
	int top = -1;
	for (int v : g[x]) if (v != fa) {
		int s = dfs(v, x);
		for (int d = 20; d > top; --d) if ((mask >> d & 1) && (s >> d & 1)) { top = d; break; }
		mask |= s;
	}
	//printf("!%d %d %d\n", x, mask, top);
	for (++top; mask >> top & 1; ++top);
	if (top > 0) mask &= ~((1 << top) - 1);
	mask |= 1 << top;
	
	//printf("%d %d %d\n", x, mask, top);

	return mask;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; ++i) {
		int u, v;
	
		scanf("%d%d", &u, &v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	int mask = dfs(1, 0), ans = 0;
	for (int i = 0; i < 20; ++i) if (mask >> i & 1) ans = i;
	printf("%d\n", ans);
	return 0;
}