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>
using namespace std;

const int N = 100 * 1000 + 5, LG = 20;
int ans = 0;

vector <int> adj[N];
int dp[N];
int cnt[N][LG];
bool mark[N];

int get(int x, int y) {
	return (x >> y) & 1;
}

void dfs(int v) {
	mark[v] = true;
	for (auto u : adj[v]) {
		if (!mark[u]) {
			dfs(u);
			dp[v] |= dp[u];
			for (int i = 0; i < LG; i++) {
				if (get(dp[u], i)) {
					cnt[v][i]++;
				}
			}
		}
	}
	int x = -1;
	for (int i = 0; i < LG; i++) {
		if (cnt[v][i] > 1) {
			x = i;
		}
	}
	x++;
	while (cnt[v][x]) {
		x++;
	}
	ans = max(ans, x);
	for (int i = 0; i < x; i++) {
		dp[v] &= (1 << LG) - 1 - (1 << i);
	}
	dp[v] |= (1 << x);
}

int main() {
	int n;
	cin >> n;
	for (int i = 1; i < n; i++) {
		int u, v;
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	dfs(1);
	cout << ans;
	return 0;
}
