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;
inline long long getint() {
  long long _x = 0, _tmp = 1;
  char _tc = getchar();
  while ((_tc < '0' || _tc > '9') && _tc != '-') _tc = getchar();
  if (_tc == '-') _tc = getchar(), _tmp = -1;
  while (_tc >= '0' && _tc <= '9') _x *= 10, _x += (_tc - '0'), _tc = getchar();
  return _x * _tmp;
}
inline long long add(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x + _y;
  if (_ >= _mod) _ -= _mod;
  return _;
}
inline long long sub(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x - _y;
  if (_ < 0) _ += _mod;
  return _;
}
inline long long mul(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x * _y;
  if (_ >= _mod) _ %= _mod;
  return _;
}
long long mypow(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 1ll;
  long long _tmp = mypow(_a, _x / 2, _mod);
  _tmp = mul(_tmp, _tmp, _mod);
  if (_x & 1) _tmp = mul(_tmp, _a, _mod);
  return _tmp;
}
long long mymul(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 0ll;
  long long _tmp = mymul(_a, _x / 2, _mod);
  _tmp = add(_tmp, _tmp, _mod);
  if (_x & 1) _tmp = add(_tmp, _a, _mod);
  return _tmp;
}
inline bool equal(double _x, double _y) {
  return _x > _y - 1e-9 && _x < _y + 1e-9;
}
int __ = 1, _cs;
void build() {}
int r, c, v[514][514];
inline long long sqr(long long x) { return x * x; }
void init() {
  r = getint();
  c = getint();
  for (int i = 0; i < r; i++)
    for (int j = 0; j < c; j++) v[i][j] = getint();
}
inline long long dif(int r1, int r2) {
  long long cu = 0;
  for (int j = 0; j < c; j++) cu += sqr(abs(v[r1][j] - v[r2][j]));
  return cu;
}
void solve() {
  long long v1 = dif(0, r - 1);
  long long v2 = dif(r / 2, r / 2 - 1);
  puts(v1 < v2 ? "YES" : "NO");
}
int main() {
  build();
  __ = getint();
  while (__--) {
    init();
    solve();
  }
}