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 <vector>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cassert>

using namespace std;

const int N = 2e5 + 7;
const int LOG = 25;

int n;
vector<int> adj[N];
int ans;

int dfs(int u, int from) {
	int all = 0, least = 0;
	for (int v : adj[u]) {
		if (v == from) continue;
		int mask = dfs(v, u);
		for (int j = 0; j < LOG; j++) {
			if ((all >> j & 1) && (mask >> j & 1)) {
				least = max(least, j + 1);
			}
		}
		all |= mask;
	}
	assert(least + 1 < LOG);
	for (int i = least; ; i++) {
		if (~all >> i & 1) {
			ans = max(ans, i);
			all ^= (1 << i);
			for (int j = i - 1; j >= 0; j--) {
				if (all >> j & 1) {
					all ^= (1 << j);
				}
			}
			return all;
		}
	}
}

int solve() {
	ans = 0;
	dfs(rand() % n + 1, 0);
	return ans;
}

int main() {
	scanf("%d", &n);
	//n = 1e5;
	for (int i = 1; i < n; i++) {
		int u, v;
		scanf("%d %d", &u, &v);
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	int ret = n;
	for (int r = 1; r <= 1; r++) {
		ret = min(ret, solve());
	}
	printf("%d\n", ret);
}
