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;
namespace patch {
template <typename T>
std::string to_string(const T& n) {
  std::ostringstream stm;
  stm << n;
  return stm.str();
}
}  // namespace patch
long long modpow(long long base, long long exponent, long long modulus) {
  long long result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = (result * base) % modulus;
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long gcd(long long u, long long v) { return (v != 0) ? gcd(v, u % v) : u; }
long long mulmod(long long a, long long b, long long c) {
  long long x = 0, y = a % c;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x + y) % c;
    }
    y = (y * 2) % c;
    b /= 2;
  }
  return x % c;
}
int q, h, w;
short int vals[601][601];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> h >> w;
    for (int j = 0; j < h; ++j) {
      for (int k = 0; k < w; ++k) {
        cin >> vals[j][k];
      }
    }
    int ansidx = 0;
    long long delta = 0;
    for (int j = 0; j < w; ++j) delta += abs(vals[1][j] - vals[0][j]);
    long long maxx = delta;
    for (int j = 2; j < h; ++j) {
      long long temp = 0;
      for (int k = 0; k < w; ++k) {
        temp += abs(vals[j][k] - vals[j - 1][k]);
      }
      if (temp > maxx) {
        maxx = temp;
        ansidx = j;
      }
      delta = temp;
    }
    if (ansidx == h / 2 || ansidx == h / 2 - 1 || ansidx == h / 2 + 1)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}