No

yes, but I've met the third task here

In B 1000ms TL and input of 1e6 which is too much for scripting languages to read.

Let $$$e(p)$$$ be exponent of prime $$$p$$$ in $$$f(x)$$$, where $$$f(x) = 1! * 2! * 3! ... * x!$$$.

Imagine extracting out single $$$p$$$ from $$$p! , (2p)!, (3p)! ...((x/p)p)!$$$ from $$$f(x)$$$, this would contribute: $$$(x/p)(x+1) - p(x/p)((x/p) + 1)/2$$$ powers of $$$p$$$.

But still you can extract $$$p's$$$ from $$$p^{2}!, (2p^{2})!, (3p^{2})! ...((x/p^{2})p^{2})!$$$, this would contribute: $$$(x/p^{2})(x+1) - p^{2}(x/p^{2})((x/p^{2}) + 1)/2$$$ powers of $$$p$$$.

Again you can extract more $$$p's$$$ from $$$p^{3}!, (2p^{3})!, (3p^{3})! ...((x/p^{3})p^{3})!$$$, this would contribute: $$$(x/p^{3})(x+1) - p^{3}(x/p^{3})((x/p^{3}) + 1)/2$$$ powers of $$$p$$$.

And so on..

So, you gotta continue summing up till $$$\lceil log_{p}(1e^{9}) \rceil$$$ iterations. That's pretty small!

Complexity: $$$O(t\lceil log_{5}N \rceil)$$$, where $$$t$$$ are no. of testcases.

Answer will be the exponent of $$$5$$$ in $$$f(x)$$$. (Of course)

Highest power of a prime $$$p$$$ which divides $$$n!$$$ is $$$\frac{n - sum(n)}{p - 1}$$$. Here $$$sum(n)$$$ is the sum of digits of $$$n$$$ in base $$$p$$$. You need to find the summation of this over $$$[a,b]$$$ for $$$p = 5$$$.

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