Read problem statements in [Mandarin], [Bengali], [Russian], and [Vietnamese] as well.

Chef and Divyam are playing a game. They start with a rational number $X$, such that $0 < X < 1$.

In each turn, Chef chooses a positive real number $Y$. Then Divyam either adds or subtracts $Y$ to $X$. So $X$ is replaced with either $X+Y$ or $X-Y$. If this number becomes $0$ or $1$, then Chef wins. Otherwise, the game continues.

You are given $X$, and you have to determine whether Chef can win in a finite number of moves. More formally, does there exist a finite number $N$ so that Chef has a strategy that is guaranteed to win in at most $N$ moves?

Here, $X$ will be given in the form $\frac{A}{B}$, where $A$ and $B$ are positive integers, $A < B$ and $\gcd(A,B)=1$.

------ Input ------
The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
The first and only line of each test case contains two space-separated integers $A$ and $B$.

------ Output ------
For each test case, print a single line containing the string "Yes" if Chef can win the game in a finite number of moves and "No" otherwise (without quotes). You can print "Yes" or "No" in any case.

------ Constraints ------
$1 ≤ T ≤ 10^{6}$
$1 ≤ A < B ≤ 10^{18}$
$\gcd(A, B)=1$

------ Subtasks ------
Subtask #1 (100 points): Original constraints

----- Sample Input 1 ------
2
1 2
2 3
----- Sample Output 1 ------
Yes
No
----- explanation 1 ------
Test case 1: Chef will choose $Y=\frac{1}{2}$, so whatever Divyam does, the number will become $0$ or $1$. Hence Chef will win in a finite number of moves.

Test case 2: No matter what Chef chooses, it can be shown that Divyam can make sure Chef never wins.

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