E1. Complex Segments (Easy Version)
time limit per test
6 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

This is the easy version of the problem. In this version, the constraints on $$$n$$$ and the time limit are lower. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  • all the subsets have the same size; and
  • a pair of segments intersects if and only if the two segments are in the same subset.

You are given $$$n$$$ segments $$$[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$$$. Find the maximum size of a complex subset of these segments.

Input

Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^3$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^4$$$) — the number of segments.

The second line of each test case contains $$$n$$$ integers $$$l_1, l_2, \ldots, l_n$$$ ($$$1 \le l_i \le 2n$$$) — the left endpoints of the segments.

The third line of each test case contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$l_i \leq r_i \le 2n$$$) — the right endpoints of the segments.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^4$$$.

Output

For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Example
Input
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
Output
3
4
4
Note

In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $$$\{\{ [1, 5], [2, 4] \}, \{ [6, 9], [8, 10] \}\}$$$.

In the third test case, it is optimal to make a single group containing all the segments except the second.