Bob is studying bitwise operation, like bitwise and operation & and bitwise or operation |. Bob was given ~N~ binary strings with the length ~L~, ~a_1, a_2, \ldots, a_N~, which consist of only ~0~ and ~1~. Bob's teacher gives him another ~M~ binary strings with length ~L~, ~b_1, b_2, \ldots, b_M~. For each binary string ~b_j~, Bob's teacher asks him the number of ways to insert ~N~ bitwise operations & or | into the sequence ~0, a_1, a_2, \ldots, a_N~ so that the result of the entire expression is equal to ~b_j~. Specially, the leftmost number in the entire expression is ~0~. In other words, Bob needs to find out the number of ways to insert operations so that ~0 \: \text{?} \: a_1 \: \text{?} \: a_2 \: \ldots \: \text{?} \: a_N \: = \: b_j~ where ~\text{?}~ is either & or |. Bob cannot change the order of strings ~a_1~ to ~a_N~.

Note, the bitwise and operation & and the bitwise or operation | have the equal precedence in this problem. When evaluating the expression, Bob will evaluate it from left to right. For example, ~1 \,|\, 0 \,\& \,1 = \,(1 | 0)\, \&\, 1 = \, 1 \,\& \, 1 = 1~

Since the answer may be very huge, output the number of ways modulo ~10^9 + 7~.

Input Specification

The first line of input contains three integers ~N~, ~L~ and ~M~ (~1 \le N \le 1000~, ~1 \le L \le 5000~, ~1 \le M \le 1000~), the number of binary strings Bob has, the length of each binary string, and the number of binary string Bob's teacher gives.

Each of the following ~N~ lines contains a binary string ~a_i~ with length ~L~ which consists of only ~0~ and ~1~.

Each fo the following ~M~ lines contains a binary string ~b_i~ with length ~L~ which consists of only ~0~ and ~1~.

Output Specification

Output ~M~ lines. Each line contains an integer, the number of ways modulo ~10^9 + 7~.

Constraints

Subtask	Points	Additional constraints
~1~	~10~	~N \le 20~, ~L \le 30~ and ~M = 1~
~2~	~20~	~N \le 1000~, ~L \le 16~
~3~	~40~	~N \le 500~, ~L \le 1000~
~4~	~30~	No additional constraints.

Sample Input 1

3 3 2
001
010
111
011
111

Sample Output 1

1
4

Explanation 1

For the ~1~st query string ~011~, there is only one way ~0 \:|\: 001 \: |\: 010\: \& \:111\: = \:011 ~

For the ~2~nd query string ~111~, Bob can put any opearations for the first two postions and put bitwise or operation at the thrid position. Thus, there are four ways, which are shown as follows.

1. ~0 \:\&\: 001 \: \&\: 010\: | \:111\: = \:111 ~
2. ~0 \:\&\: 001 \: |\: 010\: | \:111\: = \:111 ~
3. ~0 \:|\: 001 \: \&\: 010\: | \:111\: = \:111 ~
4. ~0 \:|\: 001 \: |\: 010\: | \:111\: = \:111 ~

Sample Input 2

5 5 1
01110
11011
10000
01010
00100
00100

Sample Output 2

6

Sample Input 3

10 10 3
0100011011
0110100101
1100010100
0111000110
1100011110
0001110100
0001101110
0110100001
1110001010
0010011101
0110011111
1101001010
0010001001

Sample Output 3

69
0
5