There are four points: $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, and $D(x_4, y_4)$. Write a program which determines whether the line $AB$ and the line $CD$ are parallel. If those two lines are parallel, your program should prints "YES" and if not prints "NO".



Input

Input consists of several datasets. In the first line, you are given the number of datasets $n$ ($n \leq 100$). There will be $n$ lines where each line correspondgs to each dataset. Each dataset consists of eight real numbers:

$x_1$ $y_1$ $x_2$ $y_2$ $x_3$ $y_3$ $x_4$ $y_4$


You can assume that $-100 \leq x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4 \leq 100$. Each value is a real number with at most 5 digits after the decimal point.

Output

For each dataset, print "YES" or "NO" in a line.

Example

Input

2
0.0 0.0 1.0 1.0 1.0 0.0 2.0 1.0
3.0 2.0 9.0 6.0 13.0 5.0 7.0 9.0


Output

YES
NO
import java.util.*;
import java.util.regex.*;

import static java.lang.Math.*;
import static java.lang.System.out;

// AOJ
public class Main {
	private static final Scanner sc=new Scanner(System.in);
	public static void main(String[] args) {
		new Main().AOJ0021();
	}
	
	void AOJ0021(){
		int N=sc.nextInt();
		for(int I=1; I<=N; I++){
			double[] x=new double[4], y=new double[4];
			for(int i=0; i<4; i++){
				x[i]=sc.nextDouble();
				y[i]=sc.nextDouble();
			}
			if((y[0]-y[1])/(x[0]-x[1]) == (y[2]-y[3])/(x[2]-x[3]))	out.println("YES");
			else	out.println("NO");
		}
	}
	
	void AOJ0145(){
		int n=sc.nextInt();
		
	}
	
	void AOJ0141(){
		int N=sc.nextInt();
		for(int I=1; I<=N; I++){
			int n=sc.nextInt();
			char[][] c=new char[n+4][n+4];
			int[] vx={0, 1, 0, -1};
			int[] vy={-1, 0, 1, 0};
			c=solve141(2, n+1, n, c, vx, vy);
			for(int j=2; j<=n+1; j++){
				for(int i=2; i<=n+1; i++)	out.print(c[i][j]);
				out.println();
			}
			out.println();
		}
	}
	char[][] solve141(int x, int y, int n, char[][] c, int[] vx, int[] vy){
		//範囲を超えていないか
		if(x<2&&x>n+1 && y<2&&y>n+1)	return c;
		//周りが#
		//return c;
		
		c[x][y]='#';
		
		// 1周目
		if(x==2){
			if(y!=2)	solve141(x, y-1, n, c, vx, vy);
			else	solve141(x+1, y, n, c, vx, vy);
		}
		if(y==2){
			if(x!=n+1)	solve141(x+1, y, n, c, vx, vy);
			else	solve141(x, y+1, n, c, vx, vy);
		}
		if(x==n+1){
			if(y!=2)	solve141(x, y+1,n,c,vx,vy);
			else	solve141(x-1, y,n,c,vx,vy);
		}
		if(y==n+1){
			if(x!=3)	solve141(x-1,y,n,c,vx,vy);
			else	solve141(x,y-1,n,c,vx,vy);
		}
		
		// それ以降
		for(int i=0; i<4; i++){
			if(c[x+vx[i]][y+vy[i]]!='#'){
				solve141(x+vx[i], y+vy[i],n,c,vx,vy);
			}
		}

		return c;
	}
	
	
	void AOJ0139(){
		int N=sc.nextInt();
		final Pattern ap=Pattern.compile("^>'(=+)#(=+)~$");
		for(int i=1; i<=N; i++){
			String s=sc.next();
			Matcher m=ap.matcher(s);
			if(m.matches()){
				if(m.group(1).length()==m.group(2).length()){
					out.println("A");
					continue;
				}
			}else if(Pattern.compile("^>\\^(Q=)+~~$").matcher(s).matches()){
				out.println("B");
				continue;
			}
			out.println("NA");
		}
	}
	
	void AOJ0137(){
		int N=sc.nextInt();
		for(int i=1; i<=N; i++){
			long s=sc.nextLong();
			out.println("Case "+i+":");
			for(int j=0; j<10; j++){
				s*=s;	s/=100;	s%=10000;
				out.println(s);
			}
		}
	}
	
	ArrayList<Integer> Sieve2(int N){
	    ArrayList<Integer> prime = new ArrayList<Integer>();
	    boolean[] list = new boolean[N+1];
	    Arrays.fill(list, true);
	    list[1]=false;
	    for (int i=2; i<=N; i++) {
	        if(list[i]) {
	            prime.add(i);
	            for (int j=i+i; j<=N; j+=i)    list[j] = false;
	        }
	    }
	    return prime;
	}

	boolean[] Sieve(int N){
	    boolean[] list = new boolean[N+1];
	    Arrays.fill(list, true);
	    list[1]=false;
	    for(int i=2; i<=N; i++) {
	        if(list[i]) {
	            for (int j=i+i; j<=N; j+=i)     list[j] = false;
	        }
	    }
	    return list;
	}
	
	boolean isPrime(int n){
	    for(int i=2; i*i<=n; i++){
	        if(n%i==0)     return false;
	    }
	    return true;
	}
	
	int gcd(int x, int y){
		if(y==0)	return x;
		else	return gcd(y, x%y);
	}
}