def solve(A, B, P):
    return (f(B, P) - f(A, P) + is_valid(A, P)) % (10**9 + 7)


def is_valid(A, P):
    if len(A) < 6:
        a = int(A)
        return (('3' in A) or (a % 3 == 0)) and (a % P != 0)
    sumA = sum(map(int, A))
    lowerA = int(A[-5:])
    return (('3' in A) or (sumA % 3 == 0)) and (lowerA % P != 0)


def f(A, P):
    if P == 8:
        return fcore(A, P, 4, 8750, 3750)
    elif P == 80:
        return fcore(A, P, 5, 98750, 38742)
    elif P == 800:
        return fcore(A, P, 6, 998750, 353670)

def f_naive(n, P, cum=0):
    cum %= 3
    count = 0
    for i in range(n + 1):
        if i % P == 0:
            continue
        if ('3' in str(i)) or ((i + cum) % 3 == 0):
            count += 1
    return count


def fcore(A, P, d, large, small):
    mod = 10**9 + 7
    inv9 = 111111112
    inv10 = 700000005
    if len(A) <= d:
        return f_naive(int(A), P)
    head = A[:-d]
    tail = A[-d:]
    n = len(head) - 1
    pow10 = pow(10, n, mod)
    pow9 = pow(9, n, mod)
    cum0 = 0
    cum1 = 0
    has_no_3 = True
    for a in head:
        m = ord(a) - 48  # ord('0') = 48
        cum0 = (cum0 + pow10 * m) % mod
        pow10 = (pow10 * inv10) % mod
        if has_no_3:
            if m == 3:
                has_no_3 = False
            m -= m > 3
            cum1 = (cum1 + pow9 * m) % mod
            pow9 = (pow9 * inv9) % mod
    if has_no_3:
        cum = sum(map(int, head))
        return (large * cum0 - small * cum1 + f_naive(int(tail), P, cum)) % mod
    else:
        return (large * cum0 - small * cum1 + int(tail) - int(tail)//P) % mod

A, B, P = input().split()
print(solve(A, B, int(P)))