Problem:
Score : 700 points

Problem Statement

There is an H \times W grid (H vertical, W horizontal), where each square contains a lowercase English letter.
Specifically, the letter in the square at the i-th row and j-th column is equal to the j-th character in the string S_i.

Snuke can apply the following operation to this grid any number of times:

Choose two different rows and swap them. Or, choose two different columns and swap them.

Snuke wants this grid to be symmetric.
That is, for any 1 \leq i \leq H and 1 \leq j \leq W, the letter in the square at the i-th row and j-th column and the letter in the square at the (H + 1 - i)-th row and (W + 1 - j)-th column should be equal.

Determine if Snuke can achieve this objective.

Constraints

1 \leq H \leq 12

1 \leq W \leq 12

|S_i| = W

S_i consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

H W
S_1
S_2
:
S_H

Output

If Snuke can make the grid symmetric, print YES; if he cannot, print NO.

Sample Input 1

2 3
arc
rac

Sample Output 1

YES

If the second and third columns from the left are swapped, the grid becomes symmetric, as shown in the image below:

Sample Input 2

3 7
atcoder
regular
contest

Sample Output 2

NO

Sample Input 3

12 12
bimonigaloaf
faurwlkbleht
dexwimqxzxbb
lxdgyoifcxid
ydxiliocfdgx
nfoabgilamoi
ibxbdqmzxxwe
pqirylfrcrnf
wtehfkllbura
yfrnpflcrirq
wvcclwgiubrk
lkbrwgwuiccv

Sample Output 3

YES

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>



using namespace std;

char a[20][20];

int freq[500];

main()
{
    int n, m;

    cin >> n >> m;

    if (n == 1 && m == 1)
    {
        cout << "YES";
        return 0;
    }

    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> a[i][j];
            freq[a[i][j]]++;
        }
    }
    bool fl = 0;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            if (a[i][j] != a[n - i + 1][m - j + 1])
            {
                fl = 1;
            }
        }
    }
    if (!fl)
    {
        cout << "YES";
        return 0;
    }
    fl = 0;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            if (freq[a[i][j]] & 1 && n != m || freq[a[i][j]] & 1 && freq[a[i][j]] != n && n == m )
            {
                cout << "NO";
                return 0;
            }

        }
    }

    cout << "YES";


}