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.
#!/usr/bin/python
import sys

a = []

def gen (n, d):
	global a
	if n > 0:
		a += [n]
	if d > 0:
		gen (n * 10 + 4, d - 1)
		gen (n * 10 + 7, d - 1)

def go (l, r, pos, c):
	l = max (l, pos)
	r = min (r, c)
#	print l, r, c
	return max (r - l + 1, 0) * c

def fact (n):
	if n > 12:
		return 10 ** 9 + 1
	if n == 0:
		return 1
	return n * fact (n - 1)

gen (0, 9)
a.sort ()

while True:
	s = sys.stdin.readline ().strip ()
	if s == '':
		s = sys.stdin.readline ().strip ()
		if s == '':
			break
	n, k = [int (x) for x in s.split ()]
	if n <= 12:
		if k > fact (n):
			print -1
			continue
	m = max (n - 13, 0)
	d = n - m
	res = 0
	for c in a:
		if c <= m:
			res = res + 1
	b = range (n - d + 1, n + 1)
	p = range (n - d + 1, n + 1)
	q = []
	k -= 1
	for i in range (d)[::-1]:
		j = k / fact (i)
#		print k, fact (i), j
		k %= fact (i)
		q += [b[j]]
		b = b[:j] + b[j + 1:]
#		print p, q, b
	for i in range (d):
		if p[i] in a and q[i] in a:
			res += 1
	print res
