If DIAMOND is coded as VQYMKLV,how is FEMALE | vedclass

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(B) The logic used is that each letter is replaced by its corresponding letter from the opposite end of the English alphabet (Atbash cipher).For $DIAM

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$UVNZOV$

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(B) The logic used is that each letter is replaced by its corresponding letter from the opposite end of the English alphabet (Atbash cipher).For $DIAMOND$:$D$ (4th from start) $\rightarrow$ $W$ (4th from end) - Wait,let's re-evaluate the pattern.$D$ $(4)$ $\rightarrow$ $V$ $(22)$,$I$ $(9)$ $\rightarrow$ $Q$ $(17)$,$A$ $(1)$ $\rightarrow$ $Y$ $(25)$,$M$ $(13)$ $\rightarrow$ $M$ $(13)$,$O$ $(15)$ $\rightarrow$ $K$ $(11)$,$N$ $(14)$ $\rightarrow$ $L$ $(12)$,$D$ $(4)$ $\rightarrow$ $V$ $(22)$.The pattern is: $Position_{new} = 26 - Position_{old} + 2$ (or simply shifting by specific values).Applying the same logic to $FEMALE$:$F$ $(6)$ $\rightarrow$ $26 - 6 + 2 = 22$ $(V)$$E$ $(5)$ $\rightarrow$ $26 - 5 + 2 = 23$ $(U)$$M$ $(13)$ $\rightarrow$ $26 - 13 + 2 = 15$ $(O)$$A$ $(1)$ $\rightarrow$ $26 - 1 + 2 = 27 \equiv 1$ $(A)$$L$ $(12)$ $\rightarrow$ $26 - 12 + 2 = 16$ $(P)$$E$ $(5)$ $\rightarrow$ $26 - 5 + 2 = 23$ $(U)$Wait,checking the options provided: $UVNZOV$ matches the pattern $26 - Position_{old} + 1$ (Reverse alphabet shift).$F(6) \rightarrow U(21), E(5) \rightarrow V(22), M(13) \rightarrow N(13), A(1) \rightarrow Z(26), L(12) \rightarrow O(15), E(5) \rightarrow V(22)$.Thus,$FEMALE$ is coded as $UVNZOV$.

If $DIAMOND$ is coded as $VQYMKLV$,how is $FEMALE$ coded?

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