# Chinese remainder theorem large modulo

I have the following modulo congruences:

``x ≡ 0 (mod 2) x ≡ 2 (mod 5) x ≡ 21 (mod 41) x ≡ 16793129237622992703097532489897447320171386 (mod 648250901^5) ``

I know, usually these types of problem can be solved using the ChineseRemainderTheorem, i.e:

``ChineseRemainder[{0, 2}, {2, 5}, {21, 41}, {16793129237622992703097532489897447320171386, 648250901^5}] ``

But this does not work, so I wonder how to solve this in Mathematica?

The answer should be: $$x = 45349414319770996556255505100816573064904553782$$

Any ideas?