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;

static vector<int> adj[100005];
int ans = 0;

int dp(int u, int p){
	int cnt[20];
	fill(cnt,cnt+20,0);
	for(int v : adj[u]){
		if(v == p) continue;
		int mask = dp(v, u);
		for(int i = 0;i < 20;i++){
			if(((1 << i) & mask)) cnt[i]++;
		}
	}
	
	int bit = 19;
	for(;bit >= 0 && cnt[bit] <= 1;bit--) continue;
	for(bit++;cnt[bit] > 0;bit++) continue;
	ans = max(ans,bit);
	int mask = 1 << bit;
	for(int i = bit;i < 19;i++){
		if(cnt[i] > 0) mask |= (1 << i);
	}

	return mask;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int n; cin >> n;
	for(int i = 1;i < n;i++){
		int a, b; cin >> a >> b;
		adj[a].push_back(b);
		adj[b].push_back(a);
	} 
	dp(1,0);
	cout << ans;
}