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;
const int N = 605;
const int grad = 8;
int n, m, a[N][N], b[N][N];
bool u[N][N], f[N][N];
queue<pair<int, int> > q1, q2;
void can1(int px, int py, int x, int y) {
  if (x <= 0 || y <= 0 || x > n || y > m || u[x][y] ||
      abs(a[px][py] - a[x][y]) >= grad)
    return;
  q1.push(pair<int, int>(x, y));
  u[x][y] = true;
}
void bfs1(int x, int y) {
  q1.push(pair<int, int>(x, y));
  while (!q1.empty()) {
    x = q1.front().first;
    y = q1.front().second;
    q1.pop();
    can1(x, y, x - 1, y);
    can1(x, y, x, y - 1);
    can1(x, y, x + 1, y);
    can1(x, y, x, y + 1);
    can1(x, y, x - 1, y - 1);
    can1(x, y, x + 1, y + 1);
  }
}
void can2(int px, int py, int x, int y) {
  if (x <= 0 || y <= 0 || x > n || y > m || f[x][y] ||
      abs(b[px][py] - b[x][y]) >= grad)
    return;
  q2.push(pair<int, int>(x, y));
  f[x][y] = true;
}
void bfs2(int x, int y) {
  q2.push(pair<int, int>(x, y));
  while (!q2.empty()) {
    x = q2.front().first;
    y = q2.front().second;
    q2.pop();
    can2(x, y, x - 1, y);
    can2(x, y, x, y - 1);
    can2(x, y, x + 1, y);
    can2(x, y, x, y + 1);
    can2(x, y, x - 1, y - 1);
    can2(x, y, x + 1, y + 1);
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int test;
  cin >> test;
  while (test--) {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        cin >> a[i][j];
      }
    }
    memset(u, 0, sizeof(u));
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= n / 2; i++) {
      for (int j = 1; j <= m; j++) {
        b[i][j] = a[i + n / 2][j];
        b[i + n / 2][j] = a[i][j];
      }
    }
    int k1 = 0, k2 = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        if (!u[i][j]) {
          bfs1(i, j);
          k1++;
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        if (!f[i][j]) {
          bfs2(i, j);
          k2++;
        }
      }
    }
    if (k1 < k2)
      printf("NO\n");
    else
      printf("YES\n");
  }
}