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.
import math

def good(x):
    while x > 0:
        if x % 10 != 4 and x % 10 != 7:
            return False
        x //=10
    return True

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

l = 1
r = n
if n >= 15:
    l = n-14
if n <= 15 and math.factorial(n) < k:
    print(-1)
else:
    L = r - l + 1
    a = []
    for i in range(L):
        a.append(i)
    b = []
    k -= 1
    for i in range(L):
        x = k//math.factorial(L-i-1)
        y = a[x]
        b.append(y+l)
        a.remove(y)
        k -= x * math.factorial(L-i-1)
    c = []
    if 4 < l:
        c.append(4)
    if 7 < l:
        c.append(7)
    ans = 0
    while len(c) > 0:
        ans += len(c)
        cc = []
        for x in c:
            if x * 10 + 4 < l:
                cc.append(x * 10 + 4)
            if x * 10 + 7 < l:
                cc.append(x * 10 + 7)
        c = cc
    for i in range(L):
        if good(i+l) and good(b[i]):
            ans += 1
    print(ans)
    
    