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 Scala (2.11.7) (judge verdict: Accepted)

object Main extends App {
  val A1Input  = scala.io.StdIn.readLine()
  val A1Array = A1Input.split(" ").map(_.toInt).toSeq
  val A2Input  = scala.io.StdIn.readLine()
  val A2Array = A2Input.split(" ").map(_.toInt).toSeq
  val A3Input  = scala.io.StdIn.readLine()
  val A3Array = A3Input.split(" ").map(_.toInt).toSeq

  val n = scala.io.StdIn.readInt()
  var bFindArray = Vector.empty[(Int, Int)]

  (1 to n).foreach { _=>
    val b = scala.io.StdIn.readInt()
    if (A1Array.indexOf(b) != -1) {
      bFindArray :+= (1, A1Array.indexOf(b) + 1)
    }
    if (A2Array.indexOf(b) != -1) {
      bFindArray :+= (2, A2Array.indexOf(b) + 1)
    }
    if (A3Array.indexOf(b) != -1) {
      bFindArray :+= (3, A3Array.indexOf(b) + 1)
    }
  }

  if(
    bFindArray.groupBy(b => b._1).map(b => (b._1, b._2.size)).find(b => b._2 >= 3).isDefined ||
    bFindArray.groupBy(b => b._2).map(b => (b._1, b._2.size)).find(b => b._2 >= 3).isDefined ||
    (bFindArray.find(b => b == (1, 1)).isDefined && bFindArray.find(b => b == (2, 2)).isDefined && bFindArray.find(b => b == (3, 3)).isDefined) ||
    (bFindArray.find(b => b == (1, 3)).isDefined && bFindArray.find(b => b == (2, 2)).isDefined && bFindArray.find(b => b == (3, 1)).isDefined)
  ) {
    println("Yes")
  } else {
    println("No")
  }
}