题目描述

You have $n$ lights in a row. Initially, they are all off.

You are going to press these $n$ lights one by one. When you press light $i$, light $i$ will switch its state, which means it will turn on if it's off and turn off if it's on, and then for every $j$ satisfied $i|j,i< j\le n$, press light $j$ once.

For example, if $n=4$, when you press light $1$, light $1$ will turn on, and then you will press light $2,3,4$. Since you pressed light $2$, light $2$ will turn on and you will press light $4$, which will cause light 4 to turn on. After all the operations, lights $1,2,3$ will be turned on and light $4$ is still off.

You will press these $n$ lights and do the operations as mentioned above one by one. After all the operations, you want to know whether light $k$ is on or off.

You can also use the following code to understand the meaning of the problem:

void press(int x)
{
    light[x]^=1;
    for (int y=x+x;y<=n;y+=x) press(y);
}
for (int i=1;i<=n;i++) press(i);

输入格式

There are multiple testcases.

The first line contains an integer $T(1\le T\le10^5)$, which represents the number of testcases.

Each of the testcases contains two integers $n,k(1\le k\le n\le10^6)$ in a single line.

输出格式

For each testcase, if light $k$ is turned on in the end, output $\text{YES}$, otherwise output $\text{NO}$.

2
1 1
3 2

YES
NO