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;

vector <int> V[101010];
int K[101010];
int n, ans;

int dfs(int p, int r)
{
	int k, f1, f2;
	
	f1 = f2 = 0;
	
	for(int &t: V[p]){
		if(t != r){
			k = dfs(t, p);
			f2 |= (f1 & k);
			f1 |= k;
		}
	}
	
	return (f1 | K[f2]) + 1;
}

int main()
{
	int i, a, b, s, k;
	
	scanf("%d", &n);
	
	for(s=0, i=1; i<=n; i++){
		if(!(i & (i-1))) s ++;
		K[i] = (1 << s) - 1;
	}
	
	for(i=1; i<n; i++){
		scanf("%d%d", &a, &b);
		V[a].push_back(b);
		V[b].push_back(a);
	}
	
	k = dfs(1, 0);
	
	for(i=0; i<20; i++){
		if(k & (1 << i)){
			ans = i;
		}
	}
	
	printf("%d\n", ans);
	
	return 0;
}