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;
int n, x, y;
vector<int> a[100020];
int dfs(int x, int y) {
	int c = 0, d = 0, p = 0;
	for (int i: a[x]) {
		if (i != y) {
			int u = dfs(i, x);
			d |= c & u;
			c |= u;
		}
	}
	while (1 << p < d || c >> p & 1) {
		p++;
	}
	return (c >> p | 1) << p;
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; i++) {
		scanf("%d%d", &x, &y);
		a[x].push_back(y);
		a[y].push_back(x);
	}
	printf("%d\n", 31 - __builtin_clz(dfs(1, 0)));
	return 0;
}