Problem: The Human-Cow Confederation (HC2), led by Heidi, has built a base where people and cows can hide, guarded from zombie attacks. The entrance to the base is protected by an automated gate which performs a kind of a Turing test: it shows the entering creature a photograph and asks them whether the top and bottom halves of this photograph have been swapped or not. A person (or even a cow) will have no problem answering such questions; on the other hand, a zombie would just randomly smash one of the two buttons.

The creature is asked a series of such questions. If at least 75% of them are answered correctly, the gate is unlocked; otherwise, a side door opens, beneath which a huge fan is spinning...

Heidi is now building a robot army to fight the zombies, and she wants the robots to also be able to enter the base. You are tasked with programming them to distinguish the images.

<image>

The first two images from the test set. The first picture has been rearranged, but not the second.

Input

The first line of the input contains the number q of questions (1 ≤ q ≤ 220). After that, q questions follow, each of which in the format described below.

The first line of every question contains two space-separated integers h and w (1 ≤ h, w ≤ 600) – the height (number of rows) and width (number of columns) of the photograph. (Most photographs are roughly 200 × 300.) After this, h lines follow, each describing a single row of the picture. The picture is monochrome (in shades of grey). Its i-th row is described by w space-separated integers aij (j = 1, ..., w), where aij is the brightness of the corresponding pixel (0 ≤ aij < 256, where 0 is black and 255 is white).

Each picture will be either a real-life photograph, or a real-life photograph which has been broken up into two pieces and rearranged. More precisely, in the latter case, the topmost <image> rows have been moved to the bottom of the picture. It is guaranteed that h is even.

There is only a single input file to be processed, called all.in, and it is downloadable from the online judge. You are also a given another input file, called sample.in, which contains the first 20 pictures from all.in; you are provided the correct answers for sample.in in sample.out. You are also given a directory easy_bmp, which contains the first 50 input photographs in the form of .bmp image files, as well as a directory easy_sample_original_bmp, which contains the first 20 images before rearrangement. Check the notes for the download links.

Output

Your program should print q lines. The i-th line should contain your answer for the i-th question: YES if the photograph has been rearranged and NO otherwise. Your answers will be accepted if they all conform to this format and if at least 75% of them are correct.

Because the input is rather huge, feel free to process it locally and submit just your precomputed answers (i.e., a program which just prints your output for the input file all.in).

Note

The link to download all necessary files is http://assets.codeforces.com/files/690/easy_contestant_package.zip

Solution (in CPP):

Solution:
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#include <deque>
#include <cstring>
#include <functional>
#include <climits>
#include <list>
#include <ctime>
#include <complex>
#include <ext/rope>
 
#define F1(x,y,z) for(int x=y;x<z;x++)
#define F2(x,y,z) for(int x=y;x<=z;x++)
#define F3(x,y,z) for(int x=y;x>z;x--)
#define F4(x,y,z) for(int x=y;x>=z;x--)
#define mp make_pair
#define pb push_back
#define LL long long
#define co complex<double>
#define fi first
#define se second
 
#define MAX 100005
#define AMAX 1500
#define MOD 1000000007
 
#define f(c,d) ((1<<(c))*(d))
 
using namespace std;
using namespace __gnu_cxx;

int t,n,m,x[605][605];
LL ta,tb,tc;

int main(){
	#ifdef LOCAL_PROJECT
	freopen("easy_contestant_package/sample.in","r",stdin);
    #endif
	scanf("%d",&t);
	F2(tt,1,t){
		scanf("%d%d",&n,&m);
		F1(a,0,n)F1(b,0,m)scanf("%d",&x[a][b]);
		ta=tb=0;
		F1(a,0,m){
			tc=0;
			F1(b,0,3)tc+=(x[b][a]-x[n-1-b][a])*(3-b);
			ta+=(tc*tc>500);
			tc=0;
			F1(b,0,3)tc+=(x[n/2-1-b][a]-x[n/2+b][a])*(3-b);
			tb+=(tc*tc>500);
		}
		//printf("%d %d %lld %lld ",n,m,ta,tb);
		printf(tb>ta?"YES\n":"NO\n");
	}
	#ifdef LOCAL_PROJECT
	while(1){}
    #endif
	#ifdef LOCAL_PROJECT
    system("pause");
    #endif
    return 0;
}