Don't Blame Me

给定一个长度为 $n$ 的正整数数组 $a$，求有多少个非空子序列，使得这些子序列中的元素的按位与结果在二进制下恰好有 $k$ 个 $1$。答案对 $10^9+7$ 取模。 输入包括多组测试数据。第一行包含一个正整数 $t$，表示测试数据组数。依次描述每组测试数据。 每组测试数据的第一行包含两个整数 $n$ 和 $k$，分别表示数组的长度和满足条件的按位与结果二进制中有 $k$ 个 $1$。 每组测试数据的第二行包含 $n$ 个整数 $a_i$，表示数组 $a$ 中的元素。 对于每组测试数据，输出一行一个整数，表示满足条件的子序列数量。 样例输入#1解释： 有 $5$ 个元素，其中每个元素的按位与结果都为 $1$，这意味着任意三个元素的按位与结果在二进制下都有且仅有 $1$ 个 $1$，共有 $\binom{5}{3} = 10$ 种可能性，同时每个元素自身也是符合条件的子序列，所以共有 $11$ 个非空子序列。

Sadly, the problem setter couldn't think of an interesting story, thus he just asks you to solve the following problem. Given an array $ a $ consisting of $ n $ positive integers, count the number of non-empty subsequences for which the bitwise $ \mathsf{AND} $ of the elements in the subsequence has exactly $ k $ set bits in its binary representation. The answer may be large, so output it modulo $ 10^9+7 $ . Recall that the subsequence of an array $ a $ is a sequence that can be obtained from $ a $ by removing some (possibly, zero) elements. For example, $ [1, 2, 3] $ , $ [3] $ , $ [1, 3] $ are subsequences of $ [1, 2, 3] $ , but $ [3, 2] $ and $ [4, 5, 6] $ are not. Note that $ \mathsf{AND} $ represents the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

Each test contains multiple test cases. The first line contains the number of test cases $ t $ ( $ 1 \le t \le 10^4 $ ). The description of the test cases follows. The first line of each test case consists of two integers $ n $ and $ k $ ( $ 1 \leq n \leq 2 \cdot 10^5 $ , $ 0 \le k \le 6 $ ) — the length of the array and the number of set bits that the bitwise $ \mathsf{AND} $ the counted subsequences should have in their binary representation. The second line of each test case consists of $ n $ integers $ a_i $ ( $ 0 \leq a_i \leq 63 $ ) — the array $ a $ . It is guaranteed that the sum of $ n $ over all test cases doesn't exceed $ 2 \cdot 10^5 $ .

For each test case, output a single integer — the number of subsequences that have exactly $ k $ set bits in their bitwise $ \mathsf{AND} $ value's binary representation. The answer may be large, so output it modulo $ 10^9+7 $ .

6
5 1
1 1 1 1 1
4 0
0 1 2 3
5 1
5 5 7 4 2
1 2
3
12 0
0 2 0 2 0 2 0 2 0 2 0 2
10 6
63 0 63 5 5 63 63 4 12 13

31
10
10
1
4032
15