Virat loves mathematical questions. The other day he came across an interesting question which required him to find out the number of trailing zeroes for the
function. F(n) = 1^1*2^2......N^N,where N is an integer. Virat solved the problem after a few attempts. He asked the same
question from his friend Rohit to see if he can solve it. Rohit couldn’t arrive at a solution and has asked for your help. Can you help him out?

Input

The first line contains a single integer T, the number of test cases. T test cases follow.
Each line contains a single integer
N.


Output
For each test case, output a single line containing a single integer which denotes the number of trailing zeroes for the value of the function computed for 
N.

Constraints

1 <= T <= 1000
1 <= N <= 10000


Example
Input:
3
5
10
2


Output:
5
15
0
def f(n):
    count=0
    temp=1
    for i in range(5,n+1,5):
        temp=0
        t=i
        while t%5==0:
            t/=5
            temp+=1
        count+=i*temp
    return count
t=int(raw_input())
for i in range(t):
    n=int(raw_input())
    print f(n)