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.
lucky=set([])
def gen(x):
    if x>4444444444:
        return
    lucky.add(x)
    gen(x*10+4)
    gen(x*10+7)
gen(4)
gen(7)
lucky=sorted(lucky)

n,k=map(int,raw_input().split())
z=1
t=0
while z<k:
    t+=1
    z*=t
if t>n:
    print -1
    raise SystemExit()
def before(x):
    l=0
    r=len(lucky)-1
    while r-l>1:
        if lucky[(r+l)/2]>=x:
            r=(r+l)/2
        else:
            l=(r+l)/2
    if lucky[l]>=x:
        return l
    else:
        return r
def islucky(x):
    if x==0:
        return False
    while x>0:
        if not (x%10 in [4,7]):
            return False
        x/=10
    return True
res=before(n-t+1)
if t>0:
    used=[0]*t
    perm=list()
    k-=1
    while t>0:
        z/=t
        x=0
        while used[x]==1:
            x+=1
        while k>=z:
            k-=z
            x+=1
            while used[x]==1:
                x+=1
        t-=1
        used[x]=1
        perm.append(x)
    t=len(perm)
    for i in xrange(t):
        if islucky(n-t+i+1) and islucky(n-t+perm[i]+1):
            res+=1
print res
