#include <iostream>
#include <utility>

int
main(void)
{
	long t;
	std::cin >> t;

	long a, b;
	for (long i = 0; i < t; i++)
	{
		std::cin >> a >> b;

		// A pattern emerges where, if the sum of the coin stacks is
		// even, there is no way to empty them following this scheme
		// unless the sum is divisible by 3. Also, the size of
		// one stack cannot be more than double the size of the other.

		// First solution was failing on some numbers, but since it's
		// too difficult to go through the hundred thousand outputs that
		// were tested, let's just try doing this manually.
		if (a < b)
		{
			std::swap(a, b);
		}

		// No way to empty stacks without having one with leftover coins
		if (a > 2 * b)
		{
			std::cout << "NO" << std::endl;
		}
		else
		{
			// Stacks must be divisible by three to be valid, so
			// let's reduce the values to the bare minimum.
			a %= 3;
			b %= 3;
			if (a < b)
				std::swap(a, b);
			if ((a == 2 && b == 1) || (a == b && b == 0))
				std::cout << "YES" << std::endl;
			else
				std::cout << "NO" << std::endl;

		}
	}
	return 0;
}