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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret) {
  char c;
  int sgn;
  T bit = 0.1;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && c != '.' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  if (c == ' ' || c == '\n') {
    ret *= sgn;
    return 1;
  }
  while (c = getchar(), c >= '0' && c <= '9') ret += (c - '0') * bit, bit /= 10;
  ret *= sgn;
  return 1;
}
using namespace std;
int q, h, w;
int c[666][666];
int main() {
  scanff(q);
  for (int cas = int(1); cas <= int(q); cas++) {
    scanff(h);
    scanff(w);
    int pd = 0;
    for (int i = int(1); i <= int(h); i++)
      for (int j = int(1); j <= int(w); j++) scanff(c[i][j]);
    for (int i = int(2); i <= int(h); i++) {
      double cz = 0;
      for (int j = int(1); j <= int(w); j++) {
        double fm = 0;
        int fz = 0;
        for (int k = int(j - 1); k <= int(j + 1); k++) {
          if (k >= 1 && k <= w) {
            if (k != j)
              fm += (double)(abs(c[i - 1][k] - c[i][j])) * 0.3, fz += 3;
            if (k == j)
              fm += (double)(abs(c[i - 1][k] - c[i][j])) * 0.4, fz += 4;
          }
        }
        cz += fm / double(fz);
      }
      cz = cz / double(w);
      if (cz >= 6.6) {
        printf("YES\n");
        pd = 1;
        break;
      }
    }
    if (!pd) printf("NO\n");
  }
  return 0;
}