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 INF = 2 * int(1e9);
const long long INFll = 1ll * int(1e9) * int(1e9);
const int MOD = 1000000007;
template <typename T>
ostream& operator<<(ostream& out, pair<T, T>& a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T>
istream& operator>>(istream& in, pair<T, T>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<pair<T, T> > a) {
  for (int i = 0; i < a.size(); ++i) out << a[i] << endl;
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (int i = 0; i < a.size(); ++i)
    if (i == a.size() - 1)
      out << a[i];
    else
      out << a[i] << " ";
  return out;
}
struct IN {
  bool any;
  operator bool() const { return any; }
} CIN;
inline IN& operator>>(IN& in, int& x) {
  x = 0, in.any = 0;
  char c;
  while ((c = getchar()) > 0) {
    if (c < 48 || c > 57)
      if (in.any)
        return in;
      else
        continue;
    in.any = 1, x = x * 10 + c - '0';
  }
  return in;
}
int main() {
  ios_base::sync_with_stdio(0);
  int q;
  CIN >> q;
  while (q--) {
    int n, m;
    CIN >> n >> m;
    vector<vector<int> > a(n, vector<int>(m));
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j) CIN >> a[i][j];
    int sum = 0;
    for (int j = 0; j < m; ++j) sum += abs(a[n / 2][j] - a[n / 2 - 1][j]);
    if (sum / long double(m) > 17)
      cout << "YES\n";
    else
      cout << "NO\n";
    if (0) cerr << sum / long double(m) << endl;
  }
  if (0)
    cerr << fixed << setprecision(0)
         << "TIME = " << clock() / (long double)CLOCKS_PER_SEC * 1000 << "\n";
  return 0;
}