output_description: Print the number of ways to write the numbers under the conditions, modulo
  10^9 + 7.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N M
      A_1 A_2 ... A_{N}
      B_1 B_2 ... B_{M}
attempt: # -*- coding: utf-8 -*-
  
  """
  参考：https://img.atcoder.jp/keyence2019/editorial.pdf
  """
  
  import sys, re
  from collections import deque, defaultdict, Counter
  from math import sqrt, hypot, factorial, pi, sin, cos, radians
  
  if sys.version_info.minor >= 5:
      from math import gcd
  else:
      from fractions import gcd
  from heapq import heappop, heappush, heapify, heappushpop
  from bisect import bisect_left, bisect_right
  from itertools import permutations, combinations, product
  from operator import itemgetter, mul
  from copy import deepcopy
  from functools import reduce, partial
  from fractions import Fraction
  from string import ascii_lowercase, ascii_uppercase, digits
  
  
  def input():
      return sys.stdin.readline().strip()
  
  
  def list2d(a, b, c):
      return [[c] * b for i in range(a)]
  
  
  def list3d(a, b, c, d):
      return [[[d] * c for j in range(b)] for i in range(a)]
  
  
  def ceil(x, y=1):
      return int(-(-x // y))
  
  
  def round(x):
      return int((x * 2 + 1) // 2)
  
  
  def fermat(x, y, MOD):
      return x * pow(y, MOD - 2, MOD) % MOD
  
  
  def lcm(x, y):
      return (x * y) // gcd(x, y)
  
  
  def lcm_list(nums):
      return reduce(lcm, nums, 1)
  
  
  def gcd_list(nums):
      return reduce(gcd, nums, nums[0])
  
  
  def INT():
      return int(input())
  
  
  def MAP():
      return map(int, input().split())
  
  
  def LIST():
      return list(map(int, input().split()))
  
  
  sys.setrecursionlimit(10**9)
  INF = float("inf")
  MOD = 10**9 + 7
  
  N, M = MAP()
  # 二分探索用にソートしておく
  # (どの行に何が来るかの具体的な位置は今回必要ないのでソートしていい)
  A = sorted(LIST())
  B = sorted(LIST())
  MAX = N * M
  
  # 同じリストに同じ値がいたら論外
  # (各値1回ずつしか使えないのに、違う行で最大値が同じとか不可能)
  if len(A) != len(set(A)) or len(B) != len(set(B)):
      print(0)
      exit()
  
  # 1: Aにある, 2: Bにある, 3: 両方にある, 0: どちらにもない
  flags = [0] * (MAX + 1)
  for i in range(N):
      flags[A[i]] += 1
  for i in range(M):
      flags[B[i]] += 2
  
  ans = 1
  for i in range(MAX, 0, -1):
      tmp = 1
      # どちらにもない
      if flags[i] == 0:
          # Aで自分以上の数 * Bで自分以上の数 - 自分以上の数
          tmp = (N - bisect_left(A, i)) * (M - bisect_left(B, i)) - (N * M - i)
          if tmp <= 0:
              print(0)
              exit()
      # Aにある
      elif flags[i] == 1:
          # 自分の行が決まっているので、列を調べに行く
          tmp = M - bisect_left(B, i)
      # Bにある
      elif flags[i] == 2:
          tmp = N - bisect_left(A, i)
      ans = (ans * tmp) % MOD
  print(ans)
  
problem_description: Statement
  
  Consider writing each of the integers from 1 to N \times M in a grid with N
  rows and M columns, without duplicates. Takahashi thinks it is not fun enough,
  and he will write the numbers under the following conditions:
  
    * The largest among the values in the i-th row (1 \leq i \leq N) is A_i.
    * The largest among the values in the j-th column (1 \leq j \leq M) is B_j.
  
  For him, find the number of ways to write the numbers under these conditions,
  modulo 10^9 + 7.
samples: [{"input": "2 2\n    4 3\n    3 4", "output": "2\n    \n\n(A_1, A_2) = (4, 3) and (B_1, B_2) = (3, 4). In this case, there are two ways\nto write the numbers, as follows:\n\n  * 1 in (1, 1), 4 in (1, 2), 3 in (2, 1) and 2 in (2, 2).\n  * 2 in (1, 1), 4 in (1, 2), 3 in (2, 1) and 1 in (2, 2).\n\nHere, (i, j) denotes the square at the i-th row and the j-th column.\n\n* * *"}, {"input": "3 3\n    5 9 7\n    3 6 9", "output": "0\n    \n\nSince there is no way to write the numbers under the condition, 0 should be\nprinted.\n\n* * *"}, {"input": "2 2\n    4 4\n    4 4", "output": "0\n    \n\n* * *"}, {"input": "14 13\n    158 167 181 147 178 151 179 182 176 169 180 129 175 168\n    181 150 178 179 167 180 176 169 182 177 175 159 173", "output": "343772227"}]