Let $F$ be a local dyadic number field, $\mathfrak{p}$ its maximal ideal, $(*,*)_F$ its quadratic Hilbert symbol and $e$ its ramification index (i.e. $\mathfrak{p}^e$ is exact divisor of $2$). Fix an even $s\le 2e$. What is then the smallest $t\ge 0$ such that $(U_s,U_t)_F=1$ (i.e. such that for all $a\in U_s$ and $b\in U_t$, we have $(a,b)_F=1$). Here $U_s$ are the $s$-units, i.e. $U_s=1+\mathfrak{p}^s$ for $s\ge 1$, and $U_0$ are the units of $K$.

By the Local Square Theorem the elements of $U_{2e+1}$ are squares, hence such a $t$ exists. By the very last exercise in Serre's "Local Fields" we know that $t\le 2e-s$ (in fact, I do not have a proof for this and would be grateful for a complete reference).

Is always $t=2e-s$ (which is true for $\mathbb Q$), or are there fields and even $s<2e$ with $t<2e-s$?

I know that there are explicit formulas for dyadic Hilbert symbols (Vostokov/Letsko, Henniart, ...) which possibly would enable me to work out an answer to my question. However, to become comfortable with these formulas seems to be not so obvious and I would be happy for a reference or any hint to a more conceptual proof avoiding such formulas.