$\sum _{p\le x}\frac{1}{pq}$

I was given that $\sum _{p\le x}\frac{1}{p}$ = $\mathrm{log}\mathrm{log}x$+O(1).

I need to show that $\sum _{pq\le x}\frac{1}{pq}=(\mathrm{log}\mathrm{log}x{)}^{2}+O(\mathrm{log}\mathrm{log}x)$

Here we go:

Break the sum into two sums: $\sum _{p\le x}\frac{1}{p}\sum _{q\le \frac{x}{p}}\frac{1}{q}$

Using what I was given: $(\mathrm{log}\mathrm{log}x+O(1))(\mathrm{log}\mathrm{log}\frac{x}{p}+O(1))$

Log Rules: $(\mathrm{log}\mathrm{log}x+O(1))(\mathrm{log}(\mathrm{log}x-\mathrm{log}p)+O(1))$

Algebra: $\mathrm{log}\mathrm{log}x\cdot \mathrm{log}(\mathrm{log}x-\mathrm{log}p)+O(\mathrm{log}\mathrm{log}x)$

From here I am lost. Any ideas?

I was given that $\sum _{p\le x}\frac{1}{p}$ = $\mathrm{log}\mathrm{log}x$+O(1).

I need to show that $\sum _{pq\le x}\frac{1}{pq}=(\mathrm{log}\mathrm{log}x{)}^{2}+O(\mathrm{log}\mathrm{log}x)$

Here we go:

Break the sum into two sums: $\sum _{p\le x}\frac{1}{p}\sum _{q\le \frac{x}{p}}\frac{1}{q}$

Using what I was given: $(\mathrm{log}\mathrm{log}x+O(1))(\mathrm{log}\mathrm{log}\frac{x}{p}+O(1))$

Log Rules: $(\mathrm{log}\mathrm{log}x+O(1))(\mathrm{log}(\mathrm{log}x-\mathrm{log}p)+O(1))$

Algebra: $\mathrm{log}\mathrm{log}x\cdot \mathrm{log}(\mathrm{log}x-\mathrm{log}p)+O(\mathrm{log}\mathrm{log}x)$

From here I am lost. Any ideas?