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 MAXN = 100005;
long long mod = 1000000 + 3;
long long binpow(long long a, int n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    n /= 2;
  }
  return res;
}
int a[1005][1005];
int main() {
  int q;
  scanf("%d", &q);
  for (int tt = 0; tt < q; tt++) {
    int h, w;
    scanf("%d%d", &h, &w);
    for (int i = 0; i < h; i++) {
      for (int j = 0; j < w; j++) scanf("%d", &a[i][j]);
    }
    bool bad = 0;
    for (int i = 1; i < h; i++) {
      int sum = 0, cnt1 = 0, cnt2 = 0, cnt0 = 0, more = 0;
      for (int j = 0; j < w; j++) {
        sum += abs(a[i][j] - a[i - 1][j]);
        cnt0 += abs(a[i][j] - a[i - 1][j]) < 7;
        cnt1 += abs(a[i][j] - a[i - 1][j]) < 10;
        cnt2 += abs(a[i][j] - a[i - 1][j]) < 20;
        if (abs(a[i][j] - a[i - 1][j]) < abs(a[0][j] - a[h - 1][j]) ||
            abs(a[0][j] - a[h - 1][j]) > 10)
          continue;
        else
          more++;
      }
      if ((cnt1 > 15 || cnt2 > 50) && (more < 50 || cnt1 > 50))
        continue;
      else {
        bad = 1;
        break;
      }
    }
    if (bad)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}