You are given two integers a and b. You may perform any number of operations on them (possibly zero).

During each operation you should choose any positive integer x and set a := a - x, b := b - 2x or a := a - 2x, b := b - x. Note that you may choose different values of x in different operations.

Is it possible to make a and b equal to 0 simultaneously?

Your program should answer t independent test cases.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

Then the test cases follow, each test case is represented by one line containing two integers a and b for this test case (0 ≤ a, b ≤ 10^9).

Output

For each test case print the answer to it — YES if it is possible to make a and b equal to 0 simultaneously, and NO otherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Example

Input


3
6 9
1 1
1 2


Output


YES
NO
YES

Note

In the first test case of the example two operations can be used to make both a and b equal to zero:

  1. choose x = 4 and set a := a - x, b := b - 2x. Then a = 6 - 4 = 2, b = 9 - 8 = 1; 
  2. choose x = 1 and set a := a - 2x, b := b - x. Then a = 2 - 2 = 0, b = 1 - 1 = 0. 
#_______________________________________________________________#
def fact(x):
	if x == 0:
		return 1
	else:
		return x * fact(x-1)
def lower_bound(li, num): 
	answer = -1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2
		if li[middle] >= num:
			answer = middle
			end = middle - 1
		else:
			start = middle + 1
	return answer #index where x is not less than num
def upper_bound(li, num): 
	answer = -1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2

		if li[middle] <= num:
			answer = middle
			start = middle + 1
		
		else:
			end = middle - 1
	return answer #index where x is not greater than num

def abs(x):
	return x if x >=0 else -x
def binary_search(li, val, lb, ub): 
	ans = 0
	while(lb <= ub):
		mid = (lb+ub)//2
		#print(mid, li[mid])
		if li[mid] > val:
			ub = mid-1
		elif val > li[mid]:
			lb = mid + 1
		else:
			ans = 1
			break
	return ans
def kadane(x): #maximum sum contiguous subarray
	sum_so_far = 0
	current_sum = 0
	for i in x:
		current_sum += i
		if current_sum < 0:
			current_sum = 0
		else:
			sum_so_far = mpos(sum_so_far,current_sum)
	return sum_so_far
def pref(li):
	pref_sum = [0]
	for i in li:
		pref_sum.append(pref_sum[-1] + i)
	return pref_sum
def graph(n,m):
	adj = dict()
	for i in range(1,n+1):
		adj.setdefault(i,0)
	for i in range(m):
		a,b = map(int,input().split())
		adj[a] += 1
		adj[b] += 1
	return adj
#_______________________________________________________________#
'''
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'''
import sys
import threading
from math import *
sys.setrecursionlimit(300000)
threading.stack_size(10**8)

#def main():
for _ in range(int(input())):
#for _ in range(1):
	#n = int(input())
	a,b = map(int,input().split())
	#a = list(input())
	#b = list(input())
	#x = list(map(int,input().split()))
	#b = list(map(int,input().split()))
	#adj = graph(n,m)
	if (a+b)%3 == 0 and 2*min(a,b) >= max(a,b):
		print('yes') 
	else:
		print('no')















'''		
t = threading.Thread(target=main)
t.start()
t.join()
'''









	




