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.
digits = [4, 7]
nums = [4, 7]

LIM = 10**10
while nums[-1] < LIM:
    nums = list(set(nums + [num*10 + digit for digit in digits for num in nums]))

lucky_nums = list(sorted(nums))

def e(n, k):
    fact = [1, 1]
    x = 2
    while True:
        new = fact[x-1] * x
        if new > LIM:
            break
        fact.append(new)
        x += 1

    if n < len(fact) and fact[n] < k:
        return -1

    i = 1
    while k > fact[i]:
        i += 1
    ii = i
    nums = []
    k -= 1
    start = n - i + 1
    used = [False] * 30

    while ii > 0:
        f = fact[ii-1]
        digit = k / f
        digit_2 = digit
        for j in range(len(used)):
            if not used[j]:
                if digit_2 == 0:
                    used[j] = True
                    nums.append(j + start)
                digit_2 -= 1
        k %= f
        ii -= 1

    answer_count = 0
    for num in lucky_nums:
        if num <= n:
            if num < start:
                answer_count += 1
            elif nums[num - start] in lucky_nums:
                answer_count += 1
    return answer_count

'''
print e(2, 5)
print '--'
print e(7, 3)
print '--'
print e(7, 4)
print '--'
print e(4, 1)
print '--'
print e(4, 6)
print '--'
print e(4, 7)
print '--'
'''

n, k = [int(_) for _ in raw_input().split(" ")]
print e(n, k)
