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 val[100005];
vector<int> adj[100005];
void DFS(int v, int p){
	val[v] = 0;
	for (int j: adj[v]){
		if (j != p){
			DFS(j, v);
			int k = val[j];
			for (int i = 0; i < 20; i++){
				if (val[v] & (1<<i)){
					if (k&(1<<i)){
						val[v] |= ((1<<(i+1)) - 1);
					}
				}
				else{
					val[v] |= (k&(1<<i));
				}
			}
		}
	}
	val[v]++;
}

int main(){
	int N; cin >> N;
	for (int i = 0; i < N - 1; i++){
		int u, v; cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	DFS(1, 0);
	int lg = 0;
	while (val[1]/2){
		val[1] /= 2;
		lg++;
	}
	cout << lg << endl;
}
