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<cstdio>
#include<algorithm>
using namespace std;
int n, m, X[15], Y[15], v[15];
long long S1[15], S2[15];
char p[15][15];
bool Pos() {
	int i, j;
	for (i = 1; i <= n; i++) {
		S1[i] = S2[i] = 0;
		v[i] = 0;
		long long t = 1;
		for (j = 1; j <= m; j++) {
			S1[i] = S1[i] * 26 + p[i][Y[j]] - 'a';
			S2[i] += (p[i][Y[j]] - 'a')*t;
			t *= 26;
		}
	}
	int ck = n % 2;
	for (i = 1; i <= n; i++) {
		if (v[i])continue;
		for (j = 1; j <= n; j++)if (i!=j && !v[j] && S1[i] == S2[j]) {
			v[i] = v[j] = 1;
			break;
		}
		if (j == n + 1) {
			if (S1[i] == S2[i] && ck) {
				ck--;
				continue;
			}
			return false;
		}
	}
	return true;
}
int main() {
	int i, j;
	scanf("%d%d", &n, &m);
	for (i = 1; i <= n; i++) {
		scanf("%s", p[i] + 1);
	}
	for (i = 0; i < (1 << m); i++) {
		int c = 0;
		for (j = 0; j < m; j++) {
			if ((i >> j) & 1)c++;
		}
		if (c != m / 2)continue;
		int pv = m, p1 = 1;
		for (j = 0; j < m; j++) {
			if ((i >> j) & 1)Y[pv--] = j + 1;
			else Y[p1++] = j + 1;
		}
		do {
			if (Pos()) {
				puts("YES");
				return 0;
			}
		} while (next_permutation(Y + 1, Y + pv + 1));
	}
	puts("NO");
}