Count the number of trailing 0s in the factorial of a given number.Input FormatThe first line of input contains T - the number of test cases. It is followed by T lines, each containing an integer N.Output FormatFor each test case, print the number of trailing 0s in N!, separated by a new line.Constraints1 <= T <= 100001 <= N <= 1015ExampleInput325100Output0124Explanation2! = 2 [No trailing zeros]5! = 120 [Trailing zeros=1]

The problem is asking to find the number of trailing zeros in the factorial of a given number.

Here are the steps to solve this problem:

1. Initialize a variable, say count to 0. This will store the count of trailing zeros.

2. Run a loop from 5 to N (inclusive). For each iteration, divide N by the current number and add the result to count.

3. After the loop ends, count will hold the number of trailing zeros in the factorial of N.

Here is a Python solution for the problem:

T = int(input())
for _ in range(T):
    N = int(input())
    count = 0
    i = 5
    while (N/i>=1):
        count += int(N/i)
        i *= 5
    print(count)

This solution works because a trailing zero is produced by a pair of 2 and 5 in the prime factorization of N!. Since the number of 2s is always more than or equal to the number of 5s, we just need to count the number of 5s in the prime factors of N!.

This problem is asking you to calculate the number of trailing zeros in the factorial of a given number.

Here are the steps to solve this problem:

1. First, understand what a factorial is. The factorial of a number N is the product of all positive integers less than or equal to N. For example, the factorial of 5 (denoted as 5!) is 12345 = 120.

2. A trailing zero is formed by the product of 2 and 5. In the factorial of a number, the number of