Find the least nonnegative residue of $3^{1442}$ mod 700

So I have that $700=7\cdot2^2\cdot5^2$ and I got that $3^2\equiv1\pmod2$ so then $3^{1442}\equiv1\pmod2$ also $3^2\equiv1\pmod{2^2}$ so $3^{1442}\equiv1\pmod{2^2}$ which covers one of the divisors of $700$. Im not sure if I'm supposed to use $2$ or $2^2$ and I was able to find that $3^2\equiv-1\pmod5$ so $3^{1442}\equiv-1\pmod5$, For mod $7$ I wasn't able to come up with an answer in a way like the other two, and I'm not really sure how to do this to find the least non negative residue

Go $\pmod4$, $\pmod7$ and $\pmod{25}$. We have \begin{align} 3^2 \equiv 1\pmod4\\ 3^6 \equiv 1\pmod7\\ 3^{20} \equiv 1\pmod{25} \end{align} This gives us that \begin{align} 3^{60} \equiv 1\pmod4\\ 3^{60} \equiv 1\pmod7\\ 3^{60} \equiv 1\pmod{25} \end{align} This means $$3^{60} \equiv 1\pmod{700}$$ Note that $3^{1442} = 3^{24\cdot60+2} = \left(3^{60}\right)^{24} \cdot 3^2$. Hence, we obtain $$3^{1442} \equiv 3^2\pmod{700} \equiv9\pmod{700}$$
Since $\phi(700)=240$, therefore from Euler's theorem $$3^{240} \equiv 1 \pmod{700}$$ Now $$1442 =240(6)+2$$ Therefore $$3^{1442} \equiv 3^{240(6)} \cdot 3^{2} \equiv 9 \pmod{700}$$
Carmichael function $\lambda(700)=60$
As $(3,700)=(3,3\cdot233+1)=(3,1)=1,3^{60}\equiv1\pmod{700}$
and $1442=24\cdot60+2\equiv2\pmod{60}\implies3^{1442}\equiv3^2\pmod{700}$