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 itertools import *

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

MAXK = 1000000000

LNUMS = []
lc = [ [0], [] ]
for i in range(10):
        idx = i%2
        lc[1-idx] = []
        for v in lc[idx]:
                lc[1-idx] += [v*10+4, v*10+7]
        LNUMS += lc[1-idx]

FAC = [1]
j = 1
d = 1
while d <= MAXK:
        j += 1
        d *= j
        FAC.append(d)

visit = [0]*20
c = n

def lucky(i):
        while i:
                if i % 10 not in [4, 7]:
                        return False
                i //= 10
        return True

nn = max(0, n-len(FAC))
def nextn(i):
        d = nn
        while i:
                d += 1
                if visit[d-nn] == 0:
                        i -= 1
        visit[d-nn] = 1
        return d

if n < len(FAC) and FAC[n-1] < k:
        ans = -1
else:
        k -= 1
        ans = len(list(takewhile(lambda x: x <= nn, LNUMS)))
        for i in range(nn+1, n+1):
                ii = n-i
                if ii >= len(FAC):
                        nv = i
                else:
                        di = FAC[ii-1]          # (ii-1) could be -1
                        nv = nextn(k//di+1)
                        k %= di
                if lucky(i) and lucky(nv):
                        ans += 1

print ans

			 			  			  	 	  					    	 	