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>
namespace FastIO {
template <typename T>
inline void readInteger(T& ret) {
  char ch = getchar();
  bool flag = false;
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') {
    flag = true;
    ch = getchar();
  }
  ret = 0;
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + (ch - '0');
    ch = getchar();
  }
  if (flag) ret = -ret;
}
inline int nextInt() {
  int x;
  readInteger(x);
  return x;
}
inline long long nextLong() {
  long long x;
  readInteger(x);
  return x;
}
template <typename T>
inline void outInteger(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) {
    outInteger(x / 10);
  }
  putchar(x % 10 + '0');
}
inline void newline() { putchar('\n'); }
}  // namespace FastIO
using namespace std;
using namespace FastIO;
int H, W;
int board[1100][1100];
int main() {
  int T;
  int s1, s2;
  for (scanf("%d", &T); T--;) {
    scanf("%d%d", &H, &W);
    for (int i = 0; i < H; i++)
      for (int j = 0; j < W; j++) scanf("%d", &board[i][j]);
    s1 = s2 = 0;
    for (int j = 0; j < W; j++) {
      s1 += abs(board[H / 2 - 1][j] - board[H / 2][j]);
      s2 += abs(board[0][j] - board[H - 1][j]);
    }
    if (s1 > s2)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}