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.
F = [1];
while len(F) <= 13:
    F.append(F[-1]*len(F));

def get_perm(k,S):
    if k >= F[len(S)]:
        print -1;
        exit(0);
    P = [];
    while k:
        i = 0;
        while F[len(S)-1]*(i+1) <= k:
            i += 1;
        k -= F[len(S)-1]*i;
        P.append(S[i]);
        S.pop(i);
    P += S;
    return P;

q = [0];
lucky = [];
while len(q):
    u = q.pop(0);
    if u > 10**9: continue;
    if u: lucky.append(u);
    q.append(u*10 + 4);
    q.append(u*10 + 7);

n,k = map(int,raw_input().split());
k -= 1;
while len(lucky) and lucky[-1] > n:
    lucky.pop();

L = min(13,n);
s = n - L + 1;
e = n;
ans = 0;
while len(lucky) and lucky[0] < s:
    ans += 1;
    lucky.pop(0);


P = get_perm(k,range(s,e+1));
for i in xrange(L):
    if (i + s) in lucky and P[i] in lucky:
        ans += 1;
print ans;