If $$k_1, k_2$$ are natural numbers such that
(i) $$2^{k_1} \mid 10!$$ but $$2^{k_1 + 1} \mid 10!$$ and (ii) $$5^{k_2} \mid 10!$$ but $$5^{k_2 + 1} \mid 10!$$ then

A

$$k_1 = k_2$$

B

$$k_1 < k_2$$

C

$$k_1 > k_2$$

D

$$k_1 = k_2 + 3$$