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`.

Examples

Input

2 3
arc
rac


Output

YES


Input

3 7
atcoder
regular
contest


Output

NO


Input

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


Output

YES
#include "iostream"
#include "vector"
#include "algorithm"


using namespace std;

int H, W;
char a[15][15], pc[15], pr[15];
bool mp_r(int x);
bool mp_c(int x);

bool mp_c(int x) {
    if (x > W) {
        return mp_r(1);
    } else if (pc[x]) {
        return mp_c(x + 1);
    }

    int t = 0;
    for (int i = x; i <= W; i++) {
        if (!pc[i]) t++;
    }
    for (int i = x; i <= W; i++) {
        if (pc[i]) continue;
        pc[x] = i;
        pc[i] = x;
        if (((t & 1) || (i != x)) && mp_c(x + 1)) {
            return true;
        }
        pc[x] = 0;
        pc[i] = 0;
    }

    return false;
}

bool mp_r(int x) {
    if (x > H) {
        return true;
    } else if (pr[x]) {
        return mp_r(x + 1);
    }

    int t = 0;
    for (int i = x; i <= H; i++) {
        if (!pr[i]) t++;
    }
    for (int i = x; i <= H; i++) {
        if (pr[i]) continue;
        bool ok = true;

        for (int j = 1; j <= W; j++) {
            ok &= (a[x][j] == a[i][pc[j]]);
        }

        if (((t & 1) || (i != x)) && ok) {
            pr[x] = i;
            pr[i] = x;
            if (mp_r(x + 1)) return true;
            pr[x] = 0;
            pr[i] = 0;
        }
    }

    return false;
}

int main() {

    cin >> H >> W;
    for (int i = 1; i <= H; i++) {
        for (int j = 1; j <= W; j++) {
            cin >> a[i][j];
        }
    }
   
   if (mp_c(1)) {
       cout << "YES" << endl;
   } else {
       cout << "NO" << endl;
   }
}