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;
typedef long long lint;
typedef pair<int, int> pi;
typedef vector<int> vi;

// Source is from 2004 POI. Cave

vector<int> gph[100005];
int n;

int ret;

int f(int x, int p){
	int cnt[22] = {};
	for(auto &i : gph[x]){
		if(i == p) continue;
		int v = f(i, x);
		for(int i=0; i<20; i++){
			cnt[i] += ((v >> i) & 1);
		}
	}
	int lst = -1;
	for(int i=20; i>=0; i--){
		if(cnt[i] >= 2){
			lst = i;
			break;
		}
	}
	int mask = 0;
	for(int i=lst+1; i<20; i++){
		if(cnt[i] == 0){
			ret = max(ret, i);
			for(int j=i+1; j<20; j++){
				if(cnt[j]) mask |= (1<<j);
			}
			mask |= (1<<i);
			return mask;
		}
	}
	assert(0);
}

int main(){
	cin >> n;
	for(int i=1; i<n; i++){
		int s, e;
		scanf("%d %d",&s,&e);
		gph[s].push_back(e);
		gph[e].push_back(s);
	}
	f(1, -1);
	cout << ret <<endl;
}

