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 <bits/stdc++.h>
using namespace std;
const int N = 1 << 18;
int col[N], ans, n;
vector<int> G[N];
void DFS(int v = 1, int par = 0) {
	int mx = 0, b = 0;
	for (auto u : G[v])
		if (u ^ par) {
			DFS(u, v);
			mx |= col[v] & col[u];
			col[v] |= col[u];
		}
	while ((1 << b) <= mx || col[v] & (1 << b))
		b ++;
	ans = max(ans, 31 - __builtin_clz(col[v] = col[v] & ~((1 << b) - 1) | (1 << b)));
}
int main () {
	scanf("%d", &n);
	for (int i = 1, v, u; i < n; i ++) {
		scanf("%d%d", &v, &u);
		G[v].push_back(u);
		G[u].push_back(v);
	}
	DFS();
	printf("%d", ans);
}