题解 | #约数的个数#

from math import sqrt
from math import ceil
while True:
    try:
        times=int(input())
        list1=list(map(int,input().strip().split()))
        sum=1
        count=0
        for i in range(times):
            sum=1
            j=2
            if list1[i]==1:
                print(1)
                continue
            elif list1[i]==2:
                print(2)
                continue
            elif list1[i]==3:
                print(2)
                continue
            while j*j<=list1[i]:
                if list1[i]%j==0:
                    while list1[i]%j==0:
                        count+=1
                        list1[i]//=j
                sum*=(count+1)
                j+=1
                count=0
            if list1[i]>1:
                count+=1
                sum*=(count+1)
            print(sum)  
            count=0    
    except EOFError:
        break