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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
typedef long long ll;
void my_return(ll code) { exit(code); }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll my_exp(ll a, ll b, ll MOD = numeric_limits<ll>::max()) {
  a %= MOD;
  if (b == 0) {
    return 1;
  } else {
    return ((b % 2 == 1 ? a : 1) * my_exp(a * a, b / 2, MOD)) % MOD;
  }
}
bool is_prime(ll n) {
  if (n != 2 && n % 2 == 0) {
    return false;
  } else {
    for (int i = 3; (i * i) <= n; i += 2) {
      if (n % i == 0) return false;
    }
    return true;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  ll q;
  cin >> q;
  for (ll oh_my_god = 0; oh_my_god < q; ++oh_my_god) {
    ll h, w;
    cin >> h >> w;
    vector<vector<ll> > pic(h, vector<ll>(w));
    for (ll i = 0; i < h; ++i) {
      for (ll j = 0; j < w; ++j) {
        cin >> pic[i][j];
      }
    }
    ll error = 0;
    for (ll i = 0; i < w; ++i) {
      if (numeric_limits<ll>::max() -
              (pic[0][i] - pic[h - 1][i]) * (pic[0][i] - pic[h - 1][i]) >
          error) {
        error += (pic[0][i] - pic[h - 1][i]) * (pic[0][i] - pic[h - 1][i]);
      } else {
        error = numeric_limits<ll>::max();
        break;
      }
    }
    if (error < 241064) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
  my_return(0);
}