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 = 1e5 + 9;
vector<int>V[N];

int n, f[N];
// f[i] ±íÊ¾ÒÔiÎª¸ùµÄ×ÓÊ÷ÓÐÄÄÐ©±êºÅµÄµãÐèÒª½â¾öÆ¥ÅäÎÊÌâ 
void Dfs (int u, int fa) {
	int all = 0, s = 0;
	for (auto v : V[u]) if (v != fa) {
		Dfs(v, u);
		all |= (s & f[v]);
		s |= f[v];
	}
	all = all << 1 | 1;
	for (int i = 30; ~i; --i) if (all >> i & 1) {
		while (s >> i & 1) ++i;
		f[u] = (s ^ (s & ((1 << i) - 1))) | (1 << i);
		return ;
	}
}

int main () {
	scanf("%d", &n);
	for (int i = 1, a, b; i < n; ++i) scanf("%d%d", &a, &b), V[a].push_back(b), V[b].push_back(a);
	Dfs(1, 0);
	for (int i=30;i>=0;i--) if (f[1]>>i&1) return printf("%d\n",i),0;
	return 0;
}