Consecutive Zeros Mobius

It can easily be seen that there cannot be more than 3 consecutive integers such as $\mu (n_i)=\pm 1,\mu (n)$ being the Möbius function

However, we can find arbitrary long sequences of consecutive integers with a Möbius value of 0.

Find 9 consecutive integers $n+1,n+2,...,n+9\text{ such as }\mu (n+i)=0$
(Each of them is divisible by a distinct prime square, the primes are the smallest possible).

Answer format: n

[My timing: 10 sec]