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;

const int MAX = 100000;
int N, n;
vector<int> graph[MAX];
int label[MAX];

int dfs(int node){
	label[node] = -1;
	int used = 0, more = 0;
	for (int i = 0; i < graph[node].size(); ++i){
		int next = graph[node][i];
		if (!label[next]){
			int next_used = dfs(next);
			more |= used & next_used;
			used |= next_used;
		}
	}

	more |= more >> 1;
	more |= more >> 2;
	more |= more >> 4;
	more |= more >> 8;
	more |= more >> 16;
	++more;

	int possible = (~used) & (-more);
	label[node] = possible & (-possible);

	used |= label[node];
	used &= -label[node];
	return used;
}

int main(){
	scanf("%d", &N);
	for (int i = 1; i < N; ++i){
		int x, y;
		scanf("%d%d", &x, &y);
		x--,y--;
		graph[x].push_back(y);
		graph[y].push_back(x);
	}
	dfs(0);
	int maximum = *max_element(label, label+N), result = 0;
	while (maximum){
		++result;
		maximum >>= 1;
	}
	printf("%d\n", result-1);

	return 0;
}