If $2 \cdot 4^{2k+1} + 3^{3k+1} = 11t$ and $2 \cdot 4^{2k+3} + 3^{3k+4} = 11(pt + 3^q)$,where $k, t \in Z^{+}$,then $(p, q) =$

  • A
    $(16, 3k+1)$
  • B
    $(16, 3k+4)$
  • C
    $(32, 3k+1)$
  • D
    $(32, 3k+4)$

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