Hill encryption (devised by mathematician Lester S.\ Hill in 1929) is a technique that makes use of matrices and modular arithmetic. It is ideally used with an alphabet that has a prime number of characters, so we'll use the $37$ character alphabet $ A, B, \ldots, Z, 0, 1, \ldots, 9,$ and the space character. The steps involved are the following:

1. Replace each character in the initial text (the plaintext) with the substitution $A \rightarrow 0, B\rightarrow 1, \ldots, (space) \rightarrow 36$. If the plaintext is ATTACK AT DAWN this becomes
   $$0\ 19\ 19\ 0\ 2\ 10\ 36\ 0\ 19\ 36\ 3\ 0\ 22\ 13$$

2. Group these number into three-component vectors, padding with spaces at the end if necessary. After this step we have
   $$
   \left(
   \begin{array}{c}
   0 \\ 19 \\ 19
   \end{array}
   \right)
   \left(
   \begin{array}{c}
   0 \\ 2 \\ 10
   \end{array}
   \right)
   \left(
   \begin{array}{c}
   36 \\ 0 \\ 19
   \end{array}
   \right)
   \left(
   \begin{array}{c}
   36 \\ 3 \\ 0
   \end{array}
   \right)
   \left(
   \begin{array}{c}
   22 \\ 13 \\ 36
   \end{array}
   \right)
   $$

3. Multiply each of these vectors by a predetermined $3 \times 3$ encryption matrix using modulo $37$ arithmetic. If the encryption matrix is
   $$
   \left(
   \begin{array}{ccc}
   30 & 1 & 9 \\
   4 & 23 & 7 \\
   5 & 9 & 13
   \end{array}
   \right)
   $$
   then the first vector is transformed as follows:
   $$
   \begin{eqnarray*}
   \left(
   \begin{array}{ccc}
   30 & 1 & 9 \\
   4 & 23 & 7 \\
   5 & 9 & 13
   \end{array}
   \right)
   \left(
   \begin{array}{c}
   0 \\ 19 \\ 19
   \end{array}
   \right)
   & = &
   \left(
   \begin{array}{c}
   %(30(0) + 1(19) + 9(19)) \mod 37 \\
   %(4(0) + 23(19) + 7(19)) \mod 37 \\
   %(5(0) + 9(19) + 13(19)) \mod 37
   (30 \times 0 + 1 \times 19 + 9 \times 19) \mod 37 \\
   (4 \times 0 + 23 \times 19 + 7 \times 19) \mod 37 \\
   (5 \times 0 + 9 \times 19 + 13 \times 19) \mod 37
   \end{array}
   \right) \\
   & = &
   \left(
   \begin{array}{c}
   5 \\ 15 \\ 11
   \end{array}
   \right)
   \end{eqnarray*}
   $$

4. After multiplying all the vectors by the encryption matrix, convert the resulting values back to the $37$-character alphabet and concatenate the results to obtain the encrypted ciphertext. In our example the ciphertext is FPLSFA4SUK2W9K3.

This method can be generalized to work with any $n \times n$ encryption matrix in which case the initial plaintext is broken up into vectors of length $n$. For this problem you will be given an encryption matrix and a plaintext and must compute the corresponding ciphertext.

Input begins with a line containing a positive integer $n \leq 10$ indicating the size of the matrix and the vectors to use in the encryption. After this are $n$ lines each containing $n$ non-negative integers specifying the encryption matrix. After this is a single line containing the plaintext consisting only of characters in the $37$-character alphabet specified above.

Output the corresponding ciphertext on a single line.