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Chef is tired of solving boring math problems by hand. One of these problems is summing up the products of elements from each k-subset of the set [n]. Here, a k-subset is a subset containing exactly k elements and [n] refers to the set which contains all integers between 1 and n (inclusive). More formally:

Let’s denote this number by f(n, k). Note that f(n, 0) = 1.

Since calculating f(n, k) is too boring, Chef wants to know how the divisibility of f(n, k) by a given prime p depends on k. Specifically, for a fixed n, you should compute the number of ways to select k (0 ≤ k ≤ n) so that f(n, k) isn't divisible by p. After a while of thinking, Chef realized there might be too many ways to do that. Therefore, you should compute this number modulo 10^{9}+7.

------ Input ------ 

The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
The first and only line of each test case contains two space-separated integers n and p.

------ Output ------ 

For each test case, print a single line containing one number — the number of ways to select k, modulo 10^{9}+7.

------ Constraints ------ 

$1 ≤ T ≤ 4$
$1 ≤ n < 10^{501}$
$n does not contain leading zeroes$
$2 ≤ p ≤ 100,000$
$p is prime$

------ Subtasks ------ 

Subtask #1 (10 points): n ≤ 5,000

Subtask #2 (40 points): n ≤ 100,000

Subtask #3 (50 points): original constraints

----- Sample Input 1 ------ 
1
4 2
----- Sample Output 1 ------ 
2
----- explanation 1 ------ 
Example case 1: The values of f(4, k) are as follows:

f(4, 0) = 1
f(4, 1) = 1+2+3+4 = 10
f(4, 2) = 1·2+2·3+3·4+4·1+1·3+2·4 = 35
f(4, 3) = 1·2·3+2·3·4+1·3·4+1·2·4 = 50
f(4, 4) = 1·2·3·4 = 24

Only k = 0 and k = 2 give numbers indivisible by p = 2, so the answer is 2.

Source: codechef | Difficulty: VERY_HARD
Tags: Combinatorics

Complete search

Divide and conquer

Mathematics

Polynomials and generating functions


Solutions:
import numpy as np
mod = int(1000000000.0 + 7)

def stir(p, st, en):
	if st >= en:
		return [1, en]
	m = (st + en) // 2
	ret = np.polymul(stir(p, st, m), stir(p, m + 1, en))
	return np.mod(ret, p)
for _ in range(int(input())):
	(n, p) = map(int, input().split())
	nst = 1 if n % p == p - 1 else np.sum(stir(p, 1, n % p) != 0)
	n = (n + 1) // p
	while n > 0:
		(nst, n) = (nst * (n % p + 1) % mod, n // p)
	print(nst)


import numpy as np

def poly(n, p):
	a = np.array([[1, i] for i in range(1, n + 1)])
	while n != 1:
		b = []
		for i in range(0, n // 2):
			b.append(np.polymul(a[2 * i], a[2 * i + 1]))
			for j in range(len(b[i])):
				b[i][j] = b[i][j] % p
		if len(a) % 2 == 1:
			b.append(a[n - 1])
		n = (n + 1) // 2
		a = np.array(b)
	return a

def hola(n, p):
	ho = poly(n, p)
	hol = 0
	for i in range(0, len(ho[0])):
		if ho[0][i] % p != 0:
			hol = hol + 1
	return hol
t = int(input())
while t > 0:
	(n, p) = [int(x) for x in input().split()]
	t = t - 1
	n = n + 1
	b = n % p
	if b > 1:
		ans = hola(b - 1, p)
	else:
		ans = 1
	a = 1
	mod = 1000000007
	while n > 0:
		n = n // p
		a = a * (n % p + 1) % mod
	print(a * ans % mod)


import numpy as np
mod = 10 ** 9 + 7

def fn(alist, p):
	if len(alist) > 1:
		mid = len(alist) // 2
		lefthalf = alist[:mid]
		righthalf = alist[mid:]
		p1 = fn(lefthalf, p)
		p2 = fn(righthalf, p)
		p3 = p1 * p2
		for x in range(p3.order + 1):
			p3[x] = p3[x] % p
		return p3
	elif len(alist) == 1:
		pp = np.poly1d(alist[0])
		return pp
	else:
		alist = [1]
		return np.poly1d(alist)
t = int(input())
while t > 0:
	(n, p) = map(int, input().split())
	ans = 1
	N = (n + 1) % p
	n = (n + 1) // p
	while n > 0:
		ans = ans * ((n % p + 1) % mod) % mod
		n = n // p
	l = []
	for i in range(N):
		temp = []
		temp.append(1)
		temp.append(i)
		l.append(temp)
	mul = fn(l, p)
	mul1 = 0
	for x in range(mul.order + 1):
		if mul[x] % p != 0:
			mul1 = mul1 + 1
	print(ans * mul1 % mod)
	t = t - 1


import math
mod = 10 ** 9 + 7
from numpy import polymul

def ro(x):
	while x & x - 1:
		x = (x | x >> 1) + 1
	return max(x, 1)

def bas(a, p):
	(r1, r2) = divmod(a, p)
	k = r2
	ad = 1
	while r1 > 0:
		(r1, r2) = divmod(r1, p)
		ad = ad * (r2 + 1) % mod
	return ([k], ad)

def mul(a, b):
	fin = polymul(a, b)
	fin = [i % p for i in fin]
	return fin

def repmul(k):
	while len(k) != 1:
		temp = []
		i = 0
		while i < len(k):
			if i + 1 < len(k):
				temp.append(mul(k[i], k[i + 1]))
				i += 2
			else:
				temp.append(k[i])
				i += 1
		k = temp[:]
	return k[0]
for _ in range(int(input())):
	(nn, p) = [int(i) for i in input().split()]
	mil = []
	ck = 1
	ans = 1
	(lk, ans) = bas(nn + 1, p)
	if lk[0] == 0 or lk[0] == 1:
		print(ans)
		continue
	rg = lk[0] - 1
	nn = rg
	k = [[1, (i + 1) % p] for i in range(nn)]
	k = repmul(k)
	c = 0
	for i in k:
		if i % p != 0:
			c += 1
	ans = ans * c % mod
	print(ans)


import numpy
import sys
osszes = []
try:
	while True:
		currs_ = input().split(' ')
		currs = []
		for i in currs_:
			if len(i) == 0 or not i[0].isdigit():
				continue
			else:
				currs.append(i)
		osszes += currs
except EOFError as e:
	a = 0
szamok = [int(i) for i in osszes]
mod = 1000000000.0 + 7
t = szamok[0]
T = 0
it = 1
while T < t:
	n = szamok[it]
	p = szamok[it + 1]
	it += 2
	n += 1
	pinary = []
	tmp = n
	while tmp > 0:
		pinary.append(tmp % p)
		tmp //= p

	def szorzo(L, R):
		if L > R:
			return [0]
		if L == R:
			return [1, L]
		mid = (L + R) // 2
		(res1, res2) = (szorzo(L, mid), szorzo(mid + 1, R))
		res = numpy.polymul(res1, res2)
		for i in range(0, len(res)):
			res[i] = int(res[i]) % p
		return res
	ans = 1
	if pinary[0] > 1:
		cnt = 0
		res = szorzo(1, pinary[0] - 1)
		for i in range(0, len(res)):
			if res[i] > 0:
				cnt += 1
		ans = cnt
	for i in range(1, len(pinary)):
		ans = ans * (pinary[i] + 1) % mod
	print(int(ans))
	T += 1


from itertools import combinations

def mult_polynom(polynom, number, prime):
	for elem_idx in range(len(polynom)):
		polynom[elem_idx] *= number
	polynom.append(0)
	for elem_idx in range(len(polynom) - 1, 0, -1):
		polynom[elem_idx] += polynom[elem_idx - 1] // number
		polynom[elem_idx] %= prime
	polynom[0] %= prime

def solve_tail(tail, prime):
	ans = 0
	curr_product = [1]
	for i in range(1, tail + 1):
		mult_polynom(curr_product, i, prime)
	for el in curr_product:
		if el != 0:
			ans += 1
	return ans

def solve(number=1, prime=2):
	mod = 10 ** 9 + 7
	(number, prime) = (int(x) for x in input().split())
	p_numeric = []
	num = number
	if number % prime == prime - 1:
		number //= prime
		number += 1
		p_numeric.append(0)
	while number > 0:
		p_numeric.append(number % prime)
		number //= prime
	ans = 1
	for digit in p_numeric[1:]:
		ans *= digit + 1
		ans %= mod
	if p_numeric[0] < 10003:
		ans *= solve_tail(p_numeric[0], prime)
	elif prime - p_numeric[0] > prime // 2:
		ans *= p_numeric[0]
	else:
		ans *= p_numeric[0] + 1
	ans %= mod
	print(ans)
	return ans

def naive_solve(number, prime):
	sums = []
	for amount in range(0, number + 1):
		curr_sum = 0
		for comb in combinations(range(1, number + 1), amount):
			curr_prod = 1
			for elem in comb:
				curr_prod *= elem
			curr_sum += curr_prod
		curr_sum %= prime
		sums.append(curr_sum)
	ans = 0
	for el in sums:
		if el != 0:
			ans += 1
	return ans

def test():
	primes = [2, 3, 5, 7, 11, 13, 17, 19]
	was = []
	for number in range(1, 21):
		for prime in primes:
			naiv_ans = naive_solve(number, prime)
			ans = solve(number, prime)
			if naiv_ans != ans:
				print('WA', number, prime)
				print(ans, ' != ', naiv_ans)
				was.append((number, prime, ans, naiv_ans))
			else:
				print('OK')
	print(*was, sep='\n')
reqs_nr = int(input())
for req in range(reqs_nr):
	solve()


import sys
import numpy as np

def getints():
	return map(int, sys.stdin.readline().split())

def fftconv(l0, l1, thr=512):
	n0 = len(l0)
	n1 = len(l1)
	if n0 < thr or n1 < thr:
		return np.mod(np.convolve(l0, l1), p)
	n = max(n0, n1)
	k = 0
	while 1 << k < 2 * n - 1:
		k += 1
	N = 1 << k
	h = p // 2
	l0 = np.append((l0 + h) % p - h, np.zeros(N - n0, dtype=np.float64))
	l1 = np.append((l1 + h) % p - h, np.zeros(N - n1, dtype=np.float64))
	t0 = np.fft.rfft(l0)
	t1 = np.fft.rfft(l1)
	aa = np.fft.irfft(t0 * t1)
	return np.floor(aa[:2 * n - 1] + 0.5).astype(int) % p

def nnz(N):

	def cr(a, b):
		d = b - a
		if d == 0:
			return np.array([1])
		if d == 1:
			return np.array([a, 1])
		if d == 2:
			return np.array([a * (a + 1) % p, 2 * a + 1, 1])
		if d == 3:
			return np.array([a * (a + 1) * (a + 2) % p, (3 * a * (a + 2) + 2) % p, 3 * a + 3, 1])
		m = (a + b) // 2
		return fftconv(cr(a, m), cr(m, b))
	l0 = cr(1, N + 1)
	if N < 10:
		l0 = np.mod(l0, p)
	s = 0
	for k in range(N + 1):
		s += l0[k] != 0
	return s

def basep(n, p):
	l = []
	while n > 0:
		l.append(n % p)
		n //= p
	return l

def prod(l, M=None):
	t = 1
	for x in l:
		t *= x
		if M != None:
			t = t % M
	return t
(T,) = getints()
M = 10 ** 9 + 7
for _ in range(T):
	(n, p) = getints()
	bb = basep(n, p)
	r = 0
	while r < len(bb) and bb[r] == p - 1:
		r += 1
	if r == 0:
		b = nnz(bb[0]) * prod([x + 1 for x in bb[1:]], M)
	else:
		bb.append(0)
		bb[r] += 1
		b = prod([x + 1 for x in bb[r:]], M)
	print(b % M)


import numpy as np

def get_poly(le, ri, p):
	if le == ri:
		return np.poly1d(np.array([le, 1]).astype(np.int64))
	mi = (le + ri) // 2
	p0 = get_poly(le, mi, p)
	p1 = get_poly(mi + 1, ri, p)
	pol = np.polymul(p0, p1).c
	pol = [x % p for x in pol]
	return np.poly1d(pol)

def get_nnz(n, p):
	if n == 0:
		return 1
	pol = get_poly(1, n, p)
	return sum([x != 0 for x in pol])

def main():
	mod = 1000000000 + 7
	n_case = int(input())
	for _ in range(n_case):
		(n, p) = map(int, input().split())
		r = n % p
		n //= p
		if r == p - 1:
			r = 0
			n = n + 1
		ans = int(get_nnz(r, p))
		while n != 0:
			r = n % p
			ans = ans * (r + 1) % mod
			n //= p
		print(ans)
main()