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 sys
from math import factorial

def kth_perm(seq, index):
	result = []
	fact = factorial(len(seq))
	index %= fact
	while seq:
		fact = fact / len(seq)
		choice, index = index // fact, index % fact
		result.append(seq.pop(choice))
	return result

def inc_lucky(s, index):
	if index == -1:
		s = ['4'] + s
		return s
	if s[index] == '4':
		s[index] = '7'
		return s
	s[index] = '4'
	return inc_lucky(s, index - 1)
	
def is_lucky(s):
	for e in s:
		if e != '4' and e != '7':
			return False
	return True

n, k = map(int, raw_input().split())

seq = range(max(n - 12, 1), n + 1)

if k > factorial(len(seq)):
	print -1
	sys.exit(0)

seq = kth_perm(seq, k - 1)

r = 0
num = ['4']
while int(''.join(num)) < n - 12:
	r += 1
	num = inc_lucky(num, len(num) - 1)

for i, num in enumerate(seq):
	index = i + n - len(seq) + 1
	if is_lucky(str(index)) and is_lucky(str(num)):
		r += 1

print r
