Atcoder bit operation & Conclusion + formula derivation

D- An operation & Conclusion + Formula derivation

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Conclusion ：

1. $x+y=2(x$&$y)+(x$^$y)$
2. $(x$&$y)$&$(x$^$y)=0$

This question defines $AND$ namely &

$x+y=s,x$&$y=a$

$s=2a+$$(x$^$y)（1）$
$a$&$(x$^$y)=0（2）$

from （1）（2） obtain ：$s\ge 2a,a$&$(s-2a)=0$

#include<iostream>
using namespace std;
long long a,s,t;
int main()
{
    
	cin>>t;
	while(t--)
	{
    
		cin>>a>>s;
		if((s>=2*a)&&(((s-2*a)&a)==0))
			cout<<"Yes"<<endl;
		else		
			cout<<"No"<<endl;	
	}
	return 0;
}