In this question, we can dispart the sequence, then you will obtain $$$n$$$ pairs of $$$a_i$$$ and $$$b_i$$$. To judge whether the answer is YES or NO, we can compare the steps of making $$$a_i$$$ equals to $$$0$$$. Let the steps to be $$$k_1, k_2, \dots, k_{n-1}, k_n$$$, if $$$k_1\equiv k_2\equiv \dots \equiv k_{n - 1}\equiv k_n\pmod{3}$$$, then the answer is YES, otherwise it's NO. The reason is when $$$a_i = 0, b_i = y$$$, then after two more operations, it will be $$$a_i = 0, b_i = y$$$ again and the state of these two operations are $$$a_i = y, b_i = y$$$ and $$$a_i = y, b_i = 0$$$, so it's a cycle with a period of three.

Then, the main problem is how to calculate the value of $$$k_i\bmod 3$$$. The most violent way is to do it step by step, however it will be $$$\color{red}{Time\enspace limit\enspace exceed}$$$. Accordingly, this way is not desirable. Now, let $$$x = a_i$$$ and $$$y = b_i$$$, if $$$y \gt x$$$, then after $$$3$$$ operations, $$$b_i$$$ will equals to $$$y - 2x$$$, and after $$$6$$$ operations, $$$b_i$$$ will equals to $$$y - 4x$$$, and so on. Accordingly, we can let $$$y = y \bmod 2x$$$ and because of the period is $$$3$$$, then it won't change the value of $$$k_i \bmod 3$$$. Otherwise, we just do the violent operation. The time complexity will be about $$$O(n\log n)$$$ on account of the modulo operation. It will be much faster than before!

#include <iostream>
#include <tuple>

using namespace std;

const int N = 1e5 + 10;

int Data;
int n;
int a[N], b[N];

int main()
{
	scanf("%d", &Data);
	
	while (Data --)
	{
		scanf("%d", &n);
		
		for (int i = 1; i <= n; i ++)
			scanf("%d", &a[i]);
		for (int i = 1; i <= n; i ++)
			scanf("%d", &b[i]);
			
		int t = -1, flg = 1;
		for (int i = 1; i <= n; i ++)
		{
			if (a[i] == b[i] && a[i] == 0) continue; //If a[i] and b[i] are all zero, it won't affect the answer.
			
			int x = a[i], y = b[i], v = 0;
			while (x != 0)
			{
				y %= (2 * x); //Use a more fater way
				tie(x, y) = make_pair(y, abs(x - y)); //the violent way
				v = (v + 1) % 3; //Because of the violent way, we need to increase 1 step.
			}
			
			if (t != -1 && v != t) //This means there will be two k[i] that are dissimilar in the sense of modulo 3
			{
				puts("NO");
				flg = 0;
				break;
			}
			
			t = v;
		}
		
		if (flg) puts("YES");
	}
}