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# Count Number Of Rounds

Easy
0/40
Average time to solve is 15m

## Problem statement

Given an integer ‘N’, you need to make the maximum possible number of moves where each move consists of choosing a positive integer ‘X’ > 1, such that ‘N’ is divisible by ‘X’ and replacing ‘N’ with ‘N/X’.

When ‘N’ becomes equal to 1 and there are no more possible valid moves. You need to stop and your score is equal to the number of moves made.

Given ‘N’ is of the form a! / b! ( i.e. factorial of ‘a’ divided by factorial of ‘b’) for some positive integer ‘a’ and ‘b’ (a ≥ b).

You need to find and return the maximum possible score you can achieve.

Detailed explanation ( Input/output format, Notes, Images )
Constraints:
``````1 <= T <= 10
1 <= a <= 10^5
1 <= b <= 10^5

Time Limit : 1 sec
``````
Sample Input 1 :
``````2
3 1
4 4
``````
Sample Output 1 :
``````2
0
``````
Explanation Of Sample Input 1 :
``````For the first test case, we have, a = 3 , b = 1
Since N = ( a! / b! )  , N = ( 3! / 1! ) = 6.

One of the possible ways can be to choose X = 3.
Then N = ( N / X ) = 6 / 3 = 2.
Then we choose X = 2 and N becomes 1.
So, the maximum possible number of rounds = 2.

For the second test case, we have, a = 4, b = 4
Since N = ( a! / b! )  , N = ( 4! / 4! ) = 1.
We cannot make any move, so there are 0 rounds possible.
``````
Sample Input 2 :
``````2
6 4
5 1
``````
Sample Output 2 :
``````3
5
``````
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