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 maxn = 100100, lg = 20;
vector<int> to[maxn];
int solve(int u, int p){
	vector<int> cnt(lg, 0);
	for(int v : to[u]){
		if(v == p) continue;
		int msk = solve(v, u);
		for(int j = 0; j < lg; j++) if((msk>>j)&1) cnt[j]++;
	}
	int b = lg - 1;
	while(b >= 0 && cnt[b] <= 1) b--;
	int c = b + 1;
	while(cnt[c] == 1) c++;
	int msk = 1<<c;
	for(int j = c+1; j < lg; j++) msk |= cnt[j]<<j;
	return msk;
}
int main(){
	int n; cin >> n;
	for(int i = 0; i < n - 1; i++){
		int u, v; cin >> u >> v;
		to[u].push_back(v);
		to[v].push_back(u);
	}
	int msk = solve(1, 0);
	int b = 0;
	while(msk>>b) b++;
	cout << b - 1 << endl;
	return 0;
}

