Advertising Agency
Contest: Codeforces Round 697 (Div. 3)
Difficulty rating: 1600
Tags: combinatorics, math, sortings

Masha works in an advertising agency. In order to promote the new brand, she wants to conclude contracts with some bloggers. In total, Masha has connections of $$$n$$$ different bloggers. Blogger numbered $$$i$$$ has $$$a_i$$$ followers.

Since Masha has a limited budget, she can only sign a contract with $$$k$$$ different bloggers. Of course, Masha wants her ad to be seen by as many people as possible. Therefore, she must hire bloggers with the maximum total number of followers.

Help her, find the number of ways to select $$$k$$$ bloggers so that the total number of their followers is maximum possible. Two ways are considered different if there is at least one blogger in the first way, which is not in the second way. Masha believes that all bloggers have different followers (that is, there is no follower who would follow two different bloggers).

For example, if $$$n=4$$$, $$$k=3$$$, $$$a=[1, 3, 1, 2]$$$, then Masha has two ways to select $$$3$$$ bloggers with the maximum total number of followers:

- conclude contracts with bloggers with numbers $$$1$$$, $$$2$$$ and $$$4$$$. In this case, the number of followers will be equal to $$$a_1 + a_2 + a_4 = 6$$$.
- conclude contracts with bloggers with numbers $$$2$$$, $$$3$$$ and $$$4$$$. In this case, the number of followers will be equal to $$$a_2 + a_3 + a_4 = 6$$$.

Since the answer can be quite large, output it modulo $$$10^9+7$$$.

Input format:
The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 1000$$$) — the number of bloggers and how many of them you can sign a contract with.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le n$$$) — the number of followers of each blogger.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$.

Output format:
For each test case, on a separate line output one integer — the number of ways to select $$$k$$$ bloggers so that the total number of their followers is maximum possible.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
The test case is explained in the statements.

In the second test case, the following ways are valid:

- conclude contracts with bloggers with numbers $$$1$$$ and $$$2$$$. In this case, the number of followers will be equal to $$$a_1 + a_2 = 2$$$;
- conclude contracts with bloggers with numbers $$$1$$$ and $$$3$$$. In this case, the number of followers will be equal to $$$a_1 + a_3 = 2$$$;
- conclude contracts with bloggers with numbers $$$1$$$ and $$$4$$$. In this case, the number of followers will be equal to $$$a_1 + a_4 = 2$$$;
- conclude contracts with bloggers with numbers $$$2$$$ and $$$3$$$. In this case, the number of followers will be equal to $$$a_2 + a_3 = 2$$$;
- conclude contracts with bloggers with numbers $$$2$$$ and $$$4$$$. In this case, the number of followers will be equal to $$$a_2 + a_4 = 2$$$;
- conclude contracts with bloggers with numbers $$$3$$$ and $$$4$$$. In this case, the number of followers will be equal to $$$a_3 + a_4 = 2$$$.

In the third test case, the following ways are valid:

- concludes a contract with a blogger with the number $$$2$$$. In this case, the number of followers will be equal to $$$a_2 = 2$$$.

Example input:
3
4 3
1 3 1 2
4 2
1 1 1 1
2 1
1 2

Example output:
2
6
1

Editorial:
It is obvious that Masha will enter into agreements only with bloggers that have the most subscribers. You can sort all the bloggers and greedily select the prefix.
Let $$$x$$$ — be the minimum number of subscribers for the hired blogger. Then we must hire all the bloggers who have more subscribers. Let $$$m$$$ — the number of bloggers who have more than $$$x$$$ subscribers, $$$cnt[x]$$$ — the number of bloggers who have exactly $$$x$$$ subscribers. Then we should select $$$k-m$$$ bloggers from $$$cnt[x]$$$. The number of ways to do this is equal to the binomial coefficient of $$$cnt[x]$$$ by $$$k-m$$$.
You could calculate it by searching for the inverse element modulo. Then you could calculate the factorials and use the equality $$$\binom n k = \frac{n!}{k! \cdot (n-k)!}$$$.
Alternatively, you can use the equation $$$\binom n k = \binom {n-1} {k} + \binom {n-1} {k-1}$$$ and calculate it using dynamic programming. This method is better known as the Pascal triangle.