You are given a set of N vectors, each vector consists of K integers. Vector X is equivalent to Y (denoted X ≡ Y) iff there exist a bijection and an integer r, such that for each i in the range [0..K - 1].
For example, (1, 2, 2, 3) ≡ (22, 3, 4, 22), with r = 2 and f(22) = 2, f(3) = 3 and f(4) = 1. But (22, 3, 22, 4) is not equivalent to (1, 2, 2, 3).
How many pairwise nonequivalent vectors are there in a given set of N vectors?
First number contains T (T ≤ 10), number of test cases. Each test case consists of the following. First line consists of N and K (1 ≤ N ≤ 10000, 1 ≤ K ≤ 100). N lines follow, the i-th containing K integers describing the i-th vector. The vector values are from the range [0, 109].
Output one number: the number of different vectors.
## Input
First number contains T (T ≤ 10), number of test cases. Each test case consists of the following. First line consists of N and K (1 ≤ N ≤ 10000, 1 ≤ K ≤ 100). N lines follow, the i-th containing K integers describing the i-th vector. The vector values are from the range [0, 109].
## Output
Output one number: the number of different vectors.
[samples]
**Definitions**
Let $ T \in \mathbb{Z} $ be the number of test cases.
For each test case, let $ N, K \in \mathbb{Z} $ denote the number of vectors and their dimension, respectively.
Let $ V = \{v_1, v_2, \dots, v_N\} $ be the set of vectors, where each $ v_i \in \mathbb{Z}^K $.
**Equivalence Relation**
Two vectors $ u, v \in \mathbb{Z}^K $ are equivalent ($ u \equiv v $) if there exists a bijection $ f: \text{Range}(u) \cup \text{Range}(v) \to \mathbb{Z} $ and an integer $ r \in \mathbb{Z} $ such that for all $ i \in \{0, 1, \dots, K-1\} $:
$$
f(u_i) = v_i + r
$$
**Constraints**
1. $ 1 \le T \le 10 $
2. For each test case:
- $ 1 \le N \le 10000 $
- $ 1 \le K \le 100 $
- Each component of every vector satisfies $ 0 \le v_{i,j} \le 10^9 $
**Objective**
Compute the number of equivalence classes in $ V $ under $ \equiv $.