I see, the other guy has also done it.

A 1 was impossible, B 3 impossible and on the upper right 2 impossible and from there worked it out
If on upper right 1, then all the one on the down row become 3 impossible+ the upper middle 3 impossible which makes it 5 impossible 3s so it cannot be 1 on upper right thus has to be 3 on the upper right.

From there I tried and found right solution by chance/right away.