Problem:
Score : 200 points

Problem Statement

We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}.

The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet.

Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal.

Constraints

All values in input are integers.

1 \leq A_{i, j} \leq 100

A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2))

1 \leq N \leq 10

1 \leq b_i \leq 100

b_i \neq b_j (i \neq j)

Input

Input is given from Standard Input in the following format:

A_{1, 1} A_{1, 2} A_{1, 3}
A_{2, 1} A_{2, 2} A_{2, 3}
A_{3, 1} A_{3, 2} A_{3, 3}
N
b_1
\vdots
b_N

Output

If we will have a bingo, print Yes; otherwise, print No.

Sample Input 1

84 97 66
79 89 11
61 59 7
7
89
7
87
79
24
84
30

Sample Output 1

Yes

We will mark A_{1, 1}, A_{2, 1}, A_{2, 2}, A_{3, 3}, and complete the diagonal from the top-left to the bottom-right.

Sample Input 2

41 7 46
26 89 2
78 92 8
5
6
45
16
57
17

Sample Output 2

No

We will mark nothing.

Sample Input 3

60 88 34
92 41 43
65 73 48
10
60
43
88
11
48
73
65
41
92
34

Sample Output 3

Yes

We will mark all the squares.

Solution in OCaml (4.02.3) (judge verdict: Runtime Error)

let inp_matrix : 'a array -> int -> int -> unit = fun a m n ->
  for i = 0 to m - 1 do
    for j = 0 to n - 1 do
      a.(i).(j) <- Scanf.scanf " %d" (fun x -> x)
    done
  done

let rec inp_loop : (unit -> 'a) -> int -> 'a list = fun inp_fun n ->
  if n = 0 then []
  else
    let x = inp_fun () in
    x :: inp_loop inp_fun (n - 1)

let hole a m n k =
  for i = 0 to m - 1 do
    for j = 0 to n - 1 do
      if a.(i).(j) == k then
        a.(i).(j) <- (-1)
    done
  done

let calc_ans a =
  let ans = ref "No" in
  for i = 0 to 2 do
    if (a.(i).(0) == a.(i).(1) && a.(i).(0) == a.(i).(2)) then ans := "Yes"
  done;
  for j = 0 to 2 do
    if (a.(0).(j) == a.(1).(j) && a.(0).(j) == a.(2).(j)) then ans := "Yes"
  done;
  if (a.(0).(0) == a.(1).(1) && a.(2).(2) == a.(1).(1) || a.(0).(2) == a.(1).(1) && a.(0).(2) == a.(2).(0)) then ans := "Yes";
  !ans
  
  
let () =
  let a = Array.make_matrix 3 3 0 in
  let () = inp_matrix a 3 3 in
  let n = Scanf.sscanf (input_line stdin) "%d" (fun a -> a) in
  let l = inp_loop (fun () -> Scanf.sscanf (input_line stdin) "%d" (fun a -> a)) n in
  List.iter  (hole a 3 3) l;
  let ans = calc_ans a
  in
  print_endline ans