Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

One day Petya dreamt of a lexicographically k-th permutation of integers from 1 to n. Determine how many lucky numbers in the permutation are located on the positions whose indexes are also lucky numbers.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 109) — the number of elements in the permutation and the lexicographical number of the permutation.

Output

If the k-th permutation of numbers from 1 to n does not exist, print the single number "-1" (without the quotes). Otherwise, print the answer to the problem: the number of such indexes i, that i and ai are both lucky numbers.

Examples

Input

7 4


Output

1


Input

4 7


Output

1

Note

A permutation is an ordered set of n elements, where each integer from 1 to n occurs exactly once. The element of permutation in position with index i is denoted as ai (1 ≤ i ≤ n). Permutation a is lexicographically smaller that permutation b if there is such a i (1 ≤ i ≤ n), that ai < bi, and for any j (1 ≤ j < i) aj = bj. Let's make a list of all possible permutations of n elements and sort it in the order of lexicographical increasing. Then the lexicographically k-th permutation is the k-th element of this list of permutations.

In the first sample the permutation looks like that:

1 2 3 4 6 7 5

The only suitable position is 4.

In the second sample the permutation looks like that:

2 1 3 4

The only suitable position is 4.
import math
def isLucky(num):
	s = str(num)
	for i in s:
		if i != '4' and i != '7':
			return False
	return True

def getLucky(d):
	if d == 1:
		for i in [4, 7]:
			yield str(i)
	else:
		l = [i for i in getLucky(d-1)]
		for j in l:
			yield '4'+j
		for j in l:
			yield '7'+j

luckies = []
for d in range(1,11):
	for i in getLucky(d):
		luckies.append(int(i))

def canto(l, k):
	# find the k(th) permutation of list l, k counts from 1
	n = len(l)
	p = k - 1
	a = []
	i = n - 1
	mark = [False]*(n+1)
	while len(a) < n:
		fac = math.factorial(i)

		a.append(p/fac)
		p %= fac
		i -= 1
		j = 1
		cnt = 0
		while cnt < a[-1]+1:
			if not mark[j]:
				cnt += 1
			j += 1
		a[-1] = j-1
		mark[a[-1]] = True
	return [l[i-1] for i in a]

n, k = map(int, raw_input().split())
if n < 20 and k > math.factorial(n):
	print -1
else:
	for tail in xrange(1, 31):
		fac = math.factorial(tail)
		if fac >= k:
			break
	l = [i for i in xrange(n+1-tail, n+1)]
	p_tail = canto(l, k)
	# print 'tail',tail
	# print p_tail

	ans = 0
	for lucky in luckies:
		if lucky > n: break
		# print 'lucky is',lucky
		if lucky <= n-tail:
			ans += 1
			# print lucky
		else:
			i = p_tail.index(lucky)
			if isLucky(n-tail+1+i):
				ans += 1
				# print lucky
	print ans
