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 <bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
using namespace std;

typedef pair <int, int> pii;

char a[20][20], b[20][20];
int to[20], vis[20], vvv[20];
int n, m, last;

int check(){
	memset(vvv, 0, sizeof(vvv));
	int tot = m;
	for (int i = 0; i < m; i++){
		if (vvv[i]) continue;
		for (int j = 0; j < m; j++){
			if (vvv[j] || i == j) continue;
			int flag = 1;
			for (int k = 0; k < n; k++){
				flag &= a[k][i] == a[to[k]][j];
			}
			if (flag){
				vvv[i] = vvv[j] = 1, tot -= 2;
				break;
			}
		}
	}
	if (tot >= 2) return 0;
	if (tot == 1 && m % 2 == 0) return 0;
	if (tot == 0) return 1;
	for (int i = 0; i < m; i++){
		if (vvv[i]) continue;
		for (int j = 0; j < n; j++){
			if (a[j][i] != a[to[j]][i]) return 0;
		}
		return 1;
	}
}

int dfs(int now){
	if (last == 0) return check();
	if (vis[now]) return dfs(now + 1);
	for (int i = now + (1 - last % 2); i < n; i++){
		if (vis[i]) continue;
		vis[i] = vis[now] = 1, to[now] = i, to[i] = now, last -= 2 - (i == now);
		if (dfs(now + 1)) return 1;
		vis[i] = vis[now] = 0, to[now] = to[i] = -1, last += 2 - (i == now);
	}
	return 0;
}

int main(){
	memset(to, -1, sizeof(to));
	scanf("%d%d", &n, &m), last = n;
	for (int i = 0; i < n; i++){
		scanf("%s", a[i]);
	}
	if (dfs(0)) printf("YES\n");
	else printf("NO\n");
	return 0;
}
