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;

#define MAX 100002

int n;

vector<int> v[MAX];

set<int> dp[MAX];
int over[MAX];


inline void dfs(int b,int pr=-1){
	over[b] = -1;
	dp[b].clear();
	for (int i = 0; i < v[b].size(); i++){
		if (v[b][i] == pr)continue;
		dfs(v[b][i],b);
		if (dp[b].size() < dp[v[b][i]].size()){
			swap(dp[b], dp[v[b][i]]);
		}
		for (int i2 : dp[v[b][i]]){
			if (dp[b].count(i2)){
				over[b] = max(over[b], i2);
			}
			dp[b].insert(i2);
		}
		dp[v[b][i]].clear();
	}
	auto it = dp[b].begin();
	int ex = 0;
	while (it != dp[b].end()){
		if (ex == (*it)){
			it = dp[b].erase(it);
			ex++;
			continue;
		}
		else{
			if ((*it) <= over[b]){
				it = dp[b].erase(it);
				ex = over[b]+1;
				continue;
			}
			else{
				dp[b].insert(ex);
				return;
			}
		}
	}
	dp[b].insert(ex);
}

int main(){
	cin >> n;
	for (int i = 1; i < n; i++){
		int a, b;
		scanf("%d%d", &a, &b);
		a--;
		b--;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	dfs(0);
	auto r = (*dp[0].rbegin());
	cout << r << endl;
	return 0;
}