You mention your point has order
$$93556643250795678718734474880013829509\\196181230338248789325711173791286325820$$
which factorizes as
$$2^2\cdot 3\cdot 5\cdot 7\cdot137\cdot593\cdot 24337\cdot 25589\cdot3637793\cdot5733569\cdot106831998530025000830453\cdot1975901744727669147699767.$$
Without much computing power, we can apply the Pohlig-Hellman algorithm to obtain
$$t\bmod4, t\bmod3,t\bmod5,\ldots,t\bmod5733569.$$
Using the Chinese Remainder Theorem, we can combine all these results to get
$$t\bmod\left(4\cdot3\cdot5\cdots5733569\right),$$
i.e.
$$t\bmod443208349730265573969192476820.$$
As @Ruggero remarks, the other primes are only about 80 bits. So with $\approx 40$ bits of computing power we could also break that discrete logarithm. It takes a little more effort than the others though, so if $t < 400\cdots 0$, then you may as well only do the easy ones.