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.
The first line contains two integers *n* and *k* (1 ≤ *n*, *k* ≤ 10<sup class="upper-index">9</sup>) — the number of elements in the permutation and the lexicographical number of the permutation.
7 4
1
4 7
1
7 1
2
7 5040
1
10 1023
0
7 7477
-1
10 10000
1
3 7
-1
27 1
2
40 8544
2
47 1
4
47 8547744
3
50 1000000000
4
64 87
4
98 854555
6
100 1
6
9985 5888454
30
1 1
0
1 2
-1
2 1000000000
-1
10 1000000000
-1
20 1000000000
2
777777 1
126
777777 2
125
777474 10000
120
1000000000 1
1022
777777777 5
1021
777777777 1
1022
777477774 1
989
444747744 1000000000
554
475 88555458
8
12 855448
1
20 1000000000
2
47 99998544
2
49 1000000000
2
854459 95554455
126
77777779 1000000000
508
77 47
5
6999 85488877
22
7479 58884598
24
1000000000 1000000000
1022
7 1000
0
7 124
1
7 2048
1
7 3001
1
7 127
1
7 980
1
7 5000
0
7 4095
2
7 3856
0
7 5032
0
7 4999
0
2 3
-1
2 4
-1
7 985
1
4 25
-1
6 121
1
11 39916801
-1
29 1000000000
2
10 4589
1
10 100000
1
10 98564
1
10 1
2
10 100000009
-1
from math import factorial
from itertools import permutations, islice

# Function to check if a number is lucky
def is_lucky(n):
    return all(digit in '47' for digit in str(n))

# Function to generate the kth permutation of the sequence from 1 to n
def kth_permutation(n, k):
    # Check if kth permutation is possible
    if k > factorial(n):
        return -1

    # Compute the kth permutation
    # We use islice to efficiently get the kth permutation
    sequence = range(1, n + 1)
    perm_generator = permutations(sequence)
    kth_perm = next(islice(perm_generator, k - 1, k), None)
    
    return kth_perm

# The main function to solve the problem
def count_lucky(n, k):
    permutation = kth_permutation(n, k)
    # Check if permutation exists
    if permutation == -1:
        return -1

    count = 0
    for i, num in enumerate(permutation, start=1):
        if is_lucky(i) and is_lucky(num):
            count += 1

    return count

# Get inputs and print the result
n, k = map(int, input().split())
print(count_lucky(n, k))