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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;

const int MAXN = 100000 + 10000;

int n;
vector<int> edg[MAXN];
int f[MAXN], v[MAXN];

void dfs(int x, int fa)
{
	f[x] = 0, v[x] = 0;
	int cover = 0;
	for(int i = 0; i < edg[x].size(); i++)
		if(edg[x][i] != fa)
		{
			int y = edg[x][i];
			dfs(y, x);
			cover |= f[y] & f[x];
			f[x] |= f[y];
		}
	while((1 << v[x]) < cover || (f[x] & (1 << v[x])))
		v[x]++;
	f[x] &= ~((1 << v[x]) - 1);
	f[x] |= 1 << v[x];
}

int main()
{
	ios::sync_with_stdio(false);
	// freopen("1.in", "r", stdin);
	// freopen("1.out", "w", stdout);

	cin >> n;
	for(int i = 1; i <= n - 1; i++)
	{
		int x, y;
		cin >> x >> y;
		edg[x].push_back(y), edg[y].push_back(x);
	}
	dfs(1, 0);
	int ans = 0;
	for(int i = 1; i <= n; i++)
		ans = max(ans, v[i]);
	cout << ans << endl;

	return 0;
}
