exact upper bound

Question

Define $C =\{\aleph_\delta\mid\delta <\omega_1,\delta\text{ is a limit ordinal}\}$

Define $C^{(+)} = \{\kappa^+\mid \kappa \in C\}$

Define a sequence of lengths $\aleph_{\omega_1 + 1}$ of functions in $\prod C^{(+)}$

$\vec f = \langle f_i \mid i <\aleph_{\omega_1 + 1}\rangle$

by Recursion as follows:

1. $f_0 = 0$

2. $f_{i + 1} (\alpha) = f_i (\alpha) + 1$

3. For a a limit ordinal $\delta$ we divide into cases:

   1. If there exists a cardinal $\kappa < \aleph_{\omega_1 + 1}$ so that $cf (\delta) = \kappa^{++}$, choose a club $E_\delta\subset\delta$ with cardinality $\operatorname{cf}(\delta)$, and define $f_\delta(\alpha)=\sup\{f_i(\alpha)\mid i \in E_\delta\}$.

   2. Otherwise, let $f_\delta$ be some upper bound of $\langle f_i\mid i <\delta\rangle$.

      It is not hard to show that such a function exists in $\prod C^{(+)}$

I want to prove now that $\vec f$ has an exact upper bound. That is a function $h\colon C^{(+)}\to\mathrm{On}$ in $\prod C^{(+)}/J$ where $J$ is the ideal of bounded set in $\aleph_{\omega_1}$ such that for all regular cardinal $\kappa <\aleph _{\omega_1}$ $$\{\alpha\in C^{(+)}\mid\operatorname{cf}(h(\alpha))<\kappa\}\in J$$