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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define REP(i,j,k)     for(int i = j ; i < k ; ++i)
#define RREP(i,j,k)    for(int i = j ; i >=k ; --i)
#define A    first
#define B    second
#define pb   push_back
#define mp   make_pair
#define PII pair<int , int>
#define MEM(i,j)   memset(i , j , sizeof i)
#define ALL(i)     i.begin() , i.end()
#define DBGG(i,j)     cout << i << " " << j << endl
#define DB4(i,j,k,l)  cout << i << " " << j << " " << k << " " << l << endl
#define IOS cin.tie() , cout.sync_with_stdio(0)
#define endl "\n"
///------------------------------------------------------------
#define MAX 500
#define INF 0x3f3f3f3f

int t , n , m;
int x[MAX][MAX];
int32_t main(){
	IOS;
	cin >> t;
	REP(times , 0 , t){
		cin >> n >> m;
		REP(i , 0 , n) REP(j , 0 , m){
			cin >> x[i][j];
		}
		if(n == 1){
			cout << "YES" << endl;
			continue;
		}
		int ok = 1 , tmp = max(n / 2 - 2 , 0);
		REP(j , tmp , min(tmp + 2 , n - 1)){
			int cnt = 0;
			REP(i , 0 , m){
				cnt += abs(x[j][i] - x[j + 1][i]);
			}
			if(cnt > 20 * m) ok = 0;
		}
		cout << ((ok == 0) ? "YES" : "NO") << endl;
	}
    return 0;
}