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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
// #define int long long
// #define double long double
#define rep(i,n) for (int i=0; i<(int)(n); ++i)
#define rep1(i,n) for (int i=1; i<(int)(n); ++i)
#define repeq(i,n) for (int i=0; i<=(int)(n); ++i)
#define rep1eq(i,n) for (int i=1; i<=(int)(n); ++i)
#define rrep(i,n) for (int i=(int)(n)-1; i>=0; --i)
#define rrep1(i,n) for (int i=(int)(n)-1; i>0; --i)
#define rrepeq(i,n) for (int i=(int)(n); i>=0; --i)
#define rrep1eq(i,n) for (int i=(int)(n); i>0; --i)
#define REP(i,a,b) for (int i=(int)(a); i<=(int)(b); ++i)
#define RREP(i,a,b) for (int i=(int)(a); i>=(int)(b); --i)
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vb = vector<bool>;
template<typename T> using Graph = vector<vector<T>>;
template<typename T> using Spacial = vector<vector<vector<T>>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int MOD = 1e9+7;
const int MOD2 = 998244353;
// const double EPS = 1e-9;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
string interval[2] = {" ", "\n"}; // {" ", "\n"}

template<typename T> struct is_plural : false_type{};
template<typename T1, typename T2> struct is_plural<pair<T1, T2>> : true_type{};
template<typename T> struct is_plural<vector<T>> : true_type{};
template<typename T> struct is_plural<complex<T>> : true_type{};
 
template<typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << " " << p.second; }
template<typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto itr = vec.begin(); itr != vec.end(); ++itr) is >> *itr; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { if (vec.empty()) return os; bool pl = is_plural<T>(); os << vec.front(); for (auto itr = ++vec.begin(); itr != vec.end(); ++itr) os << interval[pl] << *itr; return os; }
 
bool CoutYN(bool a, string y = "Yes", string n = "No") { cout << (a ? y : n) << "\n"; return a; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

int modpow(int a, long long n, int mod = MOD) { int ret = 1; do { if (n & 1) ret = 1LL * ret * a % mod; a = 1LL * a * a % mod; } while (n >>= 1); return ret; }

template<typename T1, typename T2> bool CompareBySecond(pair<T1, T2> a, pair<T1, T2> b) { return a.second != b.second ? a.second < b.second : a.first < b.first; }
// counterclockwise from 12 o'clock direction
template<typename T> bool CompareBySlope(pair<T, T> a, pair<T, T> b) { if (a.first * b.first < 0) return a.first < b.first; if (a.first == 0) return a.second >= 0 || b.first > 0; if (b.first == 0) return b.second < 0 && a.first < 0; return a.second * b.first < a.first * b.second; }


/* -------- <templates end> -------- */


Graph<char> g(12, vector<char>(12));

bool row(int h, int w) {
  Graph<bool> det(h, vb(h, false));

  rep(i,h) {
    set<char> self;
    rep(k,w) {
      char &c = g[i][k];
      if (self.count(c)) self.erase(c);
      else self.emplace(c);
    }
    if (self.size() < 2) {
      det[i][i] = true;
    }

    rep(j,i) {
      multiset<pair<char, char>> comb;
      rep(k,w) {
        char &c = g[i][k], &d = g[j][k];
        if (comb.count({d, c})) comb.erase(comb.find({d, c}));
        else comb.emplace(c, d);
      }
      if (comb.size() == 0 || (comb.size() == 1 && begin(comb)->first == begin(comb)->second)) {
        det[i][j] = det[j][i] = true;
      }
    }
  }

  Graph<bool> dp(h+1, vb(1<<h, false));
  dp[0][0] = true;
  rep(i,h) {
    rep(j,1<<i) {
      if (!dp[i][j]) continue;
      bitset<12> s(j);
      rep(k,i) {
        if (s[k] && det[i][k]) {
          dp[i+1][j-(1<<k)] = true;
        }
      }
      dp[i+1][j+(1<<i)] = true;
    }
  }

  if (dp[h][0]) return true;

  rep(i,h) {
    if (dp[h][1<<i] && det[i][i]) return true;
  }

  return false;
}

bool column(int h, int w) {
  rep(i,12) rep(j,i) swap(g[i][j], g[j][i]);

  return row(w, h);
}

void solve() {
  int h, w; cin >> h >> w;
  rep(i,h) rep(j,w) cin >> g[i][j];

  CoutYN(row(h, w) && column(h, w), "YES", "NO");
}


/* -------- <programs end> -------- */


signed main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(12);
  solve();
  return 0;
}