def prob315():
  MOD = 10 ** 9 + 7

  def count(A, p, e, inclusive, MOD=MOD):
    prefix = 0
    has3 = False

    f, s = 0, 0
    mid = max(0, len(A) - 4 - e)
    for i in range(mid): # xrange
      d = ord(A[i]) - 0x30
      f = (f * 10 + s) % MOD
      s = (s * 9) % MOD
      if has3:
        f = (f + d) % MOD
      else:
        s = (s + d - (d > 3)) % MOD
        f = (f + (d > 3)) % MOD
      prefix = (prefix + d) % 3
      has3 |= d == 3

    p3 = 3 * p
    curr = [0] * (4 * p)
    curr[0] = curr[1] = curr[2] = s * pow(3, MOD - 2, MOD) % MOD
    curr[p3] = f

    for i in range(mid, len(A)):
      next = [0] * (4 * p)
      d = ord(A[i]) - 0x30
      for r in range(p3):
        if not curr[r]: continue
        for j in range(10):
          k = (p3 + (r * 10 + j) % p) if j == 3 else (r * 10 + j) % p3
          next[k] = (next[k] + curr[r]) % MOD
      for r in range(p):
        if not curr[p3 + r]: continue
        for j in range(10):
          k = p3 + (r * 10 + j) % p
          next[k] = (next[k] + curr[p3 + r]) % MOD
      for j in range(d):
        k = p3 + (prefix * 10 + j) % p if (j == 3 or has3) else (prefix * 10 + j) % p3
        next[k] = (next[k] + 1) % MOD
      prefix = (prefix * 10 + d) % p3
      has3 |= d == 3
      curr = next

    if inclusive:
      k = p3 + prefix % p if has3 else prefix
      curr[k] = (curr[k] + 1) % MOD
    ret = sum(c for i, c in enumerate(curr[:p3]) if i % 3 == 0 and i % p != 0)
    ret += sum(curr[p3+1:])
    return ret % MOD

  import sys
  rl = sys.stdin.readline

  A, B, P = rl().split()
  P = int(P)

  e = 0
  while P % 10 ** (e + 1) == 0:
    e += 1

  a = count(A, P, e, False)
  b = count(B, P, e, True)
  print((b - a) % MOD)

prob315()