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 <iostream>
#include <cstdio>
#include <sstream>
#include <algorithm>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iterator>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define x0 abc
#define y0 ABC
#define x1 abcd
#define y1 ABCD
#define xn abcde
#define yn ABCDE

#define l_b lower_bound
#define u_b upper_bound

#define in insert
#define er erase

#define fi first
#define se second
#define m_p make_pair

#define p_b push_back
#define po_b pop_back

#define uns unsigned

#define ld long double
#define ll long long

#define cont continue
#define re return

#define MAXLL 9000000000000000000LL
#define MAXINT 2000000000

#define MINLL -9000000000000000000LL
#define MININT -2000000000

#define pi acos(-1.)

#define endl '\n'

#define OUT cout << "-------" << endl;

using namespace std;

using namespace __gnu_pbds;

typedef tree <int, null_type, less <int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

int a[1201][1201], n, m, i, j, t, kol;

    int Abs(int a)
    {
        re (a < 0 ? -a : a);
    }

    void solve()
    {
        cin >> n >> m;

        for (i = 1; i <= n; i++)
            for (j = 1; j <= m; j++) cin >> a[i][j];

        kol = 0;

        for (i = 1; i <= m; i++)
            if (Abs(a[n / 2][i] - a[n / 2 + 1][i]) > 120) kol++;

        if (kol > 0)
        cout << "YES";
        else cout << "NO";

        cout << endl;
    }

int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    ios::sync_with_stdio(0);
    cin.tie(0);

    cin >> t;

    while (t--) solve();

    re 0;
}