Consider the language \( L = \{ a^n b^n c^n d^m \mid n, m \ge 1 \} \). According to the Pumping Lemma, which of the following statements most accurately describes the nature of this language?

Options:
A. The language \( L \) is not a context-free language (CFL) because for some \( z = a^N b^N c^N d^M \in L \), no matter how it is split into \( z = uvwxy \), it is impossible to ensure that \( uv^i wx^i y \in L \) for all \( i \ge 0 \) when \( v \) and \( x \) contain different types of characters and \( h + f = 3 \). This contradicts the Pumping Lemma.
B. The language \( L \) is a context-free language (CFL) because \( z = a^N b^N c^N d^M \in L \) can be chosen and split into \( z = uvwxy \) in such a way that \( uv^i wx^i y \in L \) for all \( i \ge 0 \).
C. The language \( L \) is a context-free language (CFL) and can be generated by a deterministic context-free grammar, ensuring that \( a^n b^n c^n d^m \) is correctly recognized for \( n, m \ge 1 \).
D. The language \( L \) is not a context-free language (CFL) because for some \( z = a^N b^N c^N d^M \in L \), no matter how it is split into \( z = uvwxy \), it is impossible to ensure that \( uv^i wx^i y \in L \) for all \( i \ge 0 \) when \( v \) and \( x \) contain different types of characters and \( h + f \ge 1 \). This contradicts the Pumping Lemma.
E. The language \( L \) is not a context-free language (CFL) because for some \( z = a^N b^N c^N d^M \in L \), no matter how it is split into \( z = uvwxy \), it is impossible to ensure that \( uv^i wx^i y \in L \) for all \( i \ge 0 \) when \( v \) and \( x \) contain different types of characters and \( h + f < 1 \). This contradicts the Pumping Lemma.
F. The language \( L \) is not a context-free language (CFL) because no matter how \( z = a^N b^N c^N d^M \in L \) is chosen and split into \( z = uvwxy \), it is impossible to ensure that \( uv^i wx^i y \in L \) for all \( i \ge 0 \) when \( v \) and \( x \) contain different types of characters and \( h + f > 1 \). This contradicts the Pumping Lemma.
G. The language \( L \) is not a context-free language (CFL) because for some \( z = a^N b^N c^N d^M \in L \), no matter how it is split into \( z = uvwxy \), it is impossible to ensure that \( uv^i wx^i y \in L \) for all \( i \ge 0 \) when \( v \) and \( x \) contain different types of characters and \( h + f = 1 \). This contradicts the Pumping Lemma.
H. The language \( L \) is not a context-free language (CFL) because for some \( z = a^N b^N c^N d^M \in L \), no matter how it is split into \( z = uvwxy \), it is impossible to ensure that \( uv^i wx^i y \in L \) for all \( i \ge 0 \) when \( v \) and \( x \) contain different types of characters and \( h + f = 2 \). This contradicts the Pumping Lemma.
I. The language \( L \) can be made to satisfy the requirements of the Pumping Lemma by adjusting the choice of \( v \) and \( x \) so that in some cases \( uv^i wx^i y \) still belongs to \( L \), thus proving it is a context-free language (CFL).
J. The language \( L \) is not a context-free language (CFL) because for some \( z = a^N b^N c^N d^M \in L \), no matter how it is split into \( z = uvwxy \), it is impossible to ensure that \( uv^i wx^i y \in L \) for all \( i \ge 0 \) when \( v \) and \( x \) contain different types of characters and \( h + f \le 1 \). This contradicts the Pumping Lemma.
Answer: F