题意：求解$ans = sum_{i = 1}^{n}sum_{j = 1}^{m}d(ij)$
首先莫比乌斯函数有性质：$d(ij) = sum_{x | i}^{}sum_{y | j}^{}|gcd(x,y) == 1|$
那么，我们可以代入得：
$ans = sum_{i = 1}^{n}sum_{j = 1}^{m}sum_{x | i}^{}sum_{y | j}^{}|gcd(x,y) == 1| = ans = sum_{i = 1}^{n}sum_{j = 1}^{m}sum_{x | i}^{}sum_{y | j}^{}sum_{d | gcd(x,y)}^{}mu (d)$
枚举d得,$ans = sum_{i = 1}^{n}sum_{j = 1}^{m}sum_{x | i}^{}sum_{y | j}^{}sum_{d = 1}^{min(n,m)}mu (d) * |d | gcd(x,y)|$
移项得$ans = sum_{d = 1}^{min(n,m)}mu (d) sum_{i = 1}^{n}sum_{j = 1}^{m}sum_{x | i}^{}sum_{y | j}^{} * |d | gcd(x,y)|$
再化简得