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>
using namespace std;
int a[301][301], h, w, dp[301][301];
int used[601][601];
int dx[] = {0, 1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 0, 1, -1, 1, 1, -1, -1};
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
bool check(int x, int y) {
  if (x >= 1 && x <= h && y >= 1 && y <= w && !used[x][y])
    return true;
  else
    return false;
}
void dfs(int x, int y, int component) {
  used[x][y] = component;
  for (int i = 1; i <= 8; i++) {
    int kx = dx[i] + x, ky = dy[i] + y;
    if (check(kx, ky) && a[x][y] == a[kx][ky]) {
      dfs(kx, ky, component);
    }
  }
}
void init() {
  for (int i = 0; i <= 600; i++) {
    for (int j = 0; j <= 600; j++) {
      used[i][j] = 0;
    }
  }
}
int main() {
  int q;
  q = read();
  while (q--) {
    h = read(), w = read();
    init();
    for (int i = 1; i <= h; i++) {
      for (int j = 1; j <= w; j++) {
        a[i][j] = read();
      }
    }
    int cntfirst = 0, cntsecond = 0;
    for (int i = 1; i <= h; i++) {
      for (int j = 1; j <= w; j++) {
        if (!used[i][j]) dfs(i, j, cntfirst + 1), cntfirst++;
      }
    }
    for (int i = h / 2 + 1; i <= h; i++) {
      for (int j = 1; j <= w; j++) {
        dp[i - h / 2][j] = a[i][j];
      }
    }
    for (int i = 1; i <= h / 2; i++) {
      for (int j = 1; j <= w; j++) {
        dp[i + h / 2][j] = a[i][j];
      }
    }
    for (int i = 1; i <= h; i++) {
      for (int j = 1; j <= w; j++) {
        a[i][j] = dp[i][j];
      }
    }
    init();
    for (int i = 1; i <= h; i++) {
      for (int j = 1; j <= w; j++) {
        if (!used[i][j]) dfs(i, j, cntsecond + 1), cntsecond++;
      }
    }
    if (cntfirst <= cntsecond) {
      printf("NO\n");
    } else {
      printf("YES\n");
    }
  }
}