COMDIV - Number of common divisors

You will be given T (T<=10^6) pair of numbers. All you have to tell is the number of common divisors between two numbers in each pair.

Input

First line of input: T (Number of test cases)
In next T lines, each have one pair A B (0 < A, B <= 10^6)

Output

One integer describing number of common divisors between two numbers.

Example

Input:
3
100000 100000
12 24
747794 238336 Output: 36
6
2

Added by:Mir Wasi Ahmed
Date:2010-10-31
Time limit:0.600s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64
Resource:Own problem, used in UODA TST

hide comments
2018-10-30 08:55:54
What's the application of gcd in this problem??
2018-10-14 17:50:59
AC in 3rd go , just use fast input/output , no need to build sieve , just find factors for every test case in O(sqrt(n))

Last edit: 2018-10-14 17:51:15
2018-08-28 22:04:28
Spoiler Alert!
Find the gcd of two numbers and then, number of divisors. Optimise in both the cases along with faster i/o.

Last edit: 2018-08-28 22:04:53
2018-08-17 11:11:34
use Fast I/O using better mathematical logic , no need to use scanf printf..
2018-02-20 20:36:06
use scanf and printf , no need to use seive
2018-01-02 15:14:00
bhosadi waalo ac in one go ka kya show off krte rehte ho gand utha ke har jagah
2018-01-02 14:48:23
AC in 1 go :)
2017-12-27 16:39:58
Man , Never thought i would end up with fastest submission till date ( 0.10s )
Fast I/O + sieve_modified + binary_gcd + O( log(N) ) +factorization method for number of divisors got me this :)

Last edit: 2017-12-27 20:04:49
2017-12-19 18:25:20
use sieve first then gcd
AC in one go:-)
2017-11-02 21:22:19
50th =)
Be Careful with I/O.Cheers!
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