Over the Hill, Part 1

문제

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 $3737$ character alphabet A, B, $\dots \backslash ldots$, Z, 0, 1, $\dots \backslash ldots$, 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$\to 0\backslash rightarrow\; 0$, B$\to 1\backslash rightarrow\; 1$, $\dots \backslash ldots$, (space)$\to 36\backslash rightarrow\; 36$.  If the plaintext is ATTACK AT DAWN this becomes $$01919021036019363022130\backslash \; 19\backslash \; 19\backslash \; 0\backslash \; 2\backslash \; 10\backslash \; 36\backslash \; 0\backslash \; 19\backslash \; 36\backslash \; 3\backslash \; 0\backslash \; 22\backslash \; 13$$.
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 $3\times 33\; \backslash times\; 3$ encryption matrix using modulo $3737$ 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: \[\begin{eqnarray*} \left( \begin{array}{ccc} 30 &  1 &  9 \\ 4 & 23 &  7 \\ 5 &  9 & 13 \end{array} \right) \left( \begin{array}{c} 0 \\ 19 \\ 19 \end{array} \right) & = & \left( \begin{array}{c} (30 \times 0 + 1 \times 19 + 9 \times 19) \mod 37 \\ (4 \times 0 + 23 \times 19 + 7 \times 19) \mod 37 \\ (5 \times 0 + 9 \times 19 + 13 \times 19) \mod 37 \end{array} \right) \\ & = & \left( \begin{array}{c} 5 \\ 15 \\ 11 \end{array} \right) \end{eqnarray*}\]
4. After multiplying all the vectors by the encryption matrix, convert the resulting values back to the $3737$-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 $n\times nn\; \backslash times\; n$ encryption matrix in which case the initial plaintext is broken up into vectors of length $nn$.  For this problem you will be given an encryption matrix and a plaintext and must compute the corresponding ciphertext.

입력

Input begins with a line containing a positive integer $n\le 10n\; \backslash leq\; 10$ indicating the size of the matrix and the vectors to use in the encryption.  After this are $nn$ lines each containing $nn$ non-negative integers specifying the encryption matrix. After this is a single line containing the plaintext consisting only of characters in the $3737$-character alphabet specified above.

출력

Output the corresponding ciphertext on a single line.