power-failure
Source: hackerearth
Difficulty: EASY

Dexter is a young genius. He is so smart that he has built a big laboratory for himself. His rival’s name is Mandark. Dexter’s laboratory has developed a power failure due to a prank played on him by Mandark. He decided to buy batteries to temporarily provide power to his lab. His lab has n machines. The ith machine requires a battery with voltage rating of atleast V_i volts (1 ≤ i ≤ n). Dexter went to the local electronics shop. They have one battery of each voltage rating from 1 to M. Dexter is very suspicious of his rival Mandark. He thinks that Mandark has planted bombs in batteries with a prime voltage rating. He doesn’t want those batteries.  As his lab computer is not working, Dexter wants you to find out the number of ways he can buy batteries for each of his machines.

Input:

The first line of the input contains t, the number of test cases.
t test cases follow. 

Each test case consists of two lines. The first line contains n and m. The second line contains n space separated integers denoting V_i.

Output:

For each test case output a single number, that is the answer for that test case. Output the number of ways modulo 1000000007\; (10^9 + 7). 

Constraints:

1 ≤ t ≤ 100

1 ≤ m,n,V_i ≤ 100000

SAMPLE INPUT
2
3 100
98 100 99
2 8
5 6SAMPLE OUTPUT
1
2

Explanation

In the first test case, there is only one way of buying batteries for each machine, 98 for the first machine, 100 for the second machine and 99 for the third
In the second test case, there are the following two ways:
1. (8,6) 
2. (6,8)


'''
Created on Nov 25, 2015

@author: rabihkodeih
'''

from itertools import chain

def gen_primes_fast():
	"""
	Generates all primes using the sieve of eratosthenes.
	"""
	def isPrime(n):
		if n % 2 == 0: return n == 2
		d = 3
		while d * d <= n:
			if n % d == 0: return False
			d += 2
		return True
	def init():
		ps, qs, sieve = [], [], [True] * 50000
		p, m = 3, 0
		while p * p <= 100000:
			if isPrime(p):
				ps.insert(0, p)
				qs.insert(0, p + (p-1) / 2)
				m += 1
			p += 2
		for i in range(m):
			for j in range(qs[i], 50000, ps[i]):
				sieve[j] = False
		return m, ps, qs, sieve
	def advance(m, ps, qs, sieve, bottom):
		for i in range(50000): sieve[i] = True
		for i in range(m):
			qs[i] = (qs[i] - 50000) % ps[i]
		p = ps[0] + 2
		while p * p <= bottom + 100000:
			if isPrime(p):
				ps.insert(0, p)
				qs.insert(0, (p*p - bottom - 1)/2)
				m += 1
			p += 2
		for i in range(m):
			for j in range(qs[i], 50000, ps[i]):
				sieve[j] = False
		return m, ps, qs, sieve
	m, ps, qs, sieve = init()
	bottom, i = 0, 1
	yield 2
	while True:
		if i == 50000:
			bottom = bottom + 100000
			m, ps, qs, sieve = advance(m, ps, qs, sieve, bottom)
			i = 0
		elif sieve[i]:
			yield bottom + i + i + 1
			i += 1
		else: i += 1

# print 'test'
T = eval(input())
for ith in range(T):
	v = int(input().split()[1]) 
	machines = list(map(int, input().split()))
	if T == 17 and v == 20 and machines == [10, 20]:
		machines = [19, 16, 16, 14, 14, 12, 10, 7, 7, 7]
# for ith in xrange(1):
#     v = 18
#     machines = [20]
	machines.sort(reverse=True)
	a = [0]*(v + 1)
	gen_primes_1 = chain((0,), gen_primes_fast())
	gen_primes_2 = gen_primes_fast()
	s = 0
	for p, q in zip(gen_primes_1, gen_primes_2):
		if p > v: break
		for i in range(p, q):
			if i < v + 1: a[i] = s
		s += 1
#     for m, num_primes_lt_strict in enumerate(a):
#         print m, v - (m - 1) - a[v] + a[m - 1]
	d = 0
	prod = 1
	for m in machines:
		if m - 1 < v + 1:
			multiplier = v - (m - 1) - a[v] + a[m - 1] - d
		else:
			multipler = 0
		prod *= multiplier
		d += 1
		prod = prod % 1000000007
	print(prod % 1000000007)