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.
from math import factorial
import sys

def lucky(x):
	return str(x).count('4')+str(x).count('7')==len(str(x))

def fun(x):
	xs, result = str(x), 0
	for i in range(len(xs)):
		rest = 1<<(len(xs)-i-1)
		if int(xs[i]) == 4:
			continue
		elif int(xs[i]) == 7:
			result += rest
			continue
		elif int(xs[i]) < 4:
			return result
		elif 4 < int(xs[i]) < 7:
			return result + rest
		else:
			return result + rest * 2
	return result + 1

def solve(x):
	now, result = 9, 0
	while now < x:
		result += fun(now)
		now = now*10+9
	return result + fun(x)

n, k = [int(t) for t in raw_input().split()]
m = 1
while factorial(m) < k:
	m += 1
if m > n:
	print -1
	sys.exit(0)
lst = range(n-m+1,n+1)
for i in range(m):
	s = sorted(lst[i:])
	j = 0
	while factorial(m-i-1) < k:
		j += 1
		k -= factorial(m-i-1)
	s[0], s[j] = s[j], s[0]
	lst = lst[:i] + s

#print range(1,n-m+1) + lst
print solve(n-m) + len([1 for a, b in zip(lst, range(n-m+1,n+1)) if lucky(a) and lucky(b)])
