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;

int main() {
    int n, m;
    cin >> n >> m;
    vector<string> s(n);
    for(int i = 0; i < n; ++i) {
        cin >> s[i];
    }

    vector<int> p1(n, -1);
    function<bool(int)> dfs1 = [&](int depth) {
        if(depth == n / 2) {
            int rest = -1;
            for(int i = 0; i < n; ++i) {
                if(p1[i] == -1) rest = i;
            }
            if(rest != -1) p1[rest] = rest;

            vector<bool> dp(1 << m);
            dp[0] = true;
            for(int S = 0; S < (1 << m); ++S) {
                if(!dp[S]) continue;
                for(int i = 0; i < m; ++i) {
                    if(S & (1 << i)) continue;
                    for(int j = i + 1; j < m; ++j) {
                        bool ok = true;
                        for(int k = 0; k < n; ++k) {
                            int y1 = k, y2 = p1[k];
                            ok &= s[y1][i] == s[y2][j];
                        }
                        if(ok) {
                            dp[S | (1 << i) | (1 << j)] = true;
                        }
                    }
                }
            }
            const int mask = (1 << m) - 1;
            bool res = dp[mask];
            if(m & 1) {
                for(int i = 0; i < m; ++i) {
                    if(dp[mask ^ (1 << i)]) {
                        bool check = true;
                        for(int j = 0; j < n; ++j) {
                            int y1 = j, y2 = p1[j];
                            check &= s[y1][i] == s[y2][i];
                        }
                        res |= check;
                    }
                }
            }
            if(rest != -1) p1[rest] = -1;
            return res;
        }

        bool res = false;
        for(int i = depth; i < n; ++i) {
            if(p1[i] != -1) continue;
            for(int j = i + 1; j < n; ++j) {
                if(p1[j] != -1) continue;
                p1[i] = j;
                p1[j] = i;
                res |= dfs1(depth + 1);
                p1[i] = p1[j] = -1;
            }
        }
        return res;
    };

    cout << (dfs1(0) ? "YES" : "NO") << endl;
}