Next smallest dimension of Specht Module after $(n)$, $(1^n)$, $(n-1,1)$ and $(2,1^{n-2})$

In representation theory of $S_n$, we know that for $n \geq 9$, the only Specht modules $S^\alpha$ of dimension $f^\alpha < {n-1 \choose 2} - 1$ are:

$S^{(n)}$ and $S^{(1^n)}$ with dimension $1$, $S^{(n-1,1)}$ and $S^{(2,1^{n-2})}$ with dimension $n-1$.

By simple calculations, we know that the next smallest is of dimension ${n-1 \choose 2} - 1$ (it is calculated by considering $\alpha = (n-2,2)$ and $(2^2,1^{n-4})$). The question is:

Are $(n-2,2)$ and $(2^2,1^{n-4})$ the only Specht modules with dimension ${n-1 \choose 2} - 1$?

Further Question: The next smallest dimension after ${n-1 \choose 2} - 1$ is ${n-1 \choose 2}$. Are $(n-2,1^2)$ and $(3,1^{n-3})$ the only Specht modules with dimension ${n-1 \choose 2}$?

Thanks for any help!

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I guess your other question, which is solved, is stronger than this one. –  Harry Huang Mar 6 '13 at 15:58
but that doesn't answer to this, isn't it? Correct me if I'm wrong :) –  terrylsc Mar 6 '13 at 23:42