Let $K$ be a number field, $\mathcal{O}_K$ be its ring of integers. Let $\mathfrak{q}$ be a nonzero ideal in $\mathcal{O}_K$. 
The $\mathfrak{q}$-ideal class group consists of fractional ideals with equivalence relation:
$$
\mathfrak{a} \sim \mathfrak{b}$$
if and only if there are $\alpha, \beta \in \mathcal{O}_K$ such that $\alpha \mathfrak{a} = \beta \mathfrak{b}$, with $\alpha$, and $\beta$ totally positive. 

Let $x\geq 1$, and $[\mathfrak{a}_0]_x$ be the set of integral ideals $\mathfrak{a}$ such that $N\mathfrak{a}\leq x$ and $\mathfrak{a}\sim \mathfrak{a}_0$. Let $h(\mathfrak{q})$ be the order of $\mathfrak{q}$-ideal class group. 

I wonder if there is a good upper bound of size of $[\mathfrak{a}_0]_x$:

Suppose that $N\mathfrak{q}>\frac{x}{2}$. 

In particular, Is it true that:
$$
|[\mathfrak{a}_0]_x|\leq B$$ 
for some constant $B$ depending only on $K$? 

My thoughts on this problem is that 

1. This is true when $K=\mathbb{Q}$. 

2. Using inverse class of $\mathfrak{a}_0$, we get a constant bound but it depends on $\mathfrak{a}_0$. 

3. In Hinz and Lodemann's paper 'On Siegel Zeros of Hecke Landau Zeta Functions', it was mentioned that 
$$
|[\mathfrak{a}_0]_x|\ll \frac{x+N\mathfrak{q}}{h(\mathfrak{q})}$$
But, this is not quite enough for boundedness.