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>
using namespace std;

int h, w, ar[15];
bool used[15], ev;
char s[12][12];

inline bool isRev(char a[], char b[]) {
	for (int i = 0; i < h; i++)if (a[i] != b[h - 1 - i])return false;
	return true;
}

bool dfs(int p, int n, int e) {
	if (e) {
		for (int i = 0; i < h; i++) {
			used[i] = true; ar[h / 2] = i;
			if (dfs(0, 0, 0))return true;
			used[i] = false;
		}
		return false;
	}
	else if (n != h / 2) {
		while (used[p])p++;
		for (int i = p + 1; i < h; i++)if (!used[i]) {
			used[i] = true;
			ar[n] = p; ar[h - 1 - n] = i;
			if (dfs(p + 1, n + 1, 0))return true;
			used[i] = false;
		}
		return false;
	}
	else {
		char tmp[15][15];
		bool u[15] = {}, ev = !(w % 2);
		for (int i = 0; i < w; i++)for (int j = 0; j < h; j++)tmp[i][j] = s[ar[j]][i];
		for (int i = 0; i < w; i++){
			for (int j = i + 1; j < w && !u[i]; j++)if (!u[j]) {
				u[i] = u[j] = isRev(tmp[i], tmp[j]);
			}
			if (!u[i] && !ev)u[i] = ev = isRev(tmp[i], tmp[i]);
			if (!u[i])return false;
		}
		return true;
	}
}

int main() {
	scanf("%d%d", &h, &w);
	for (int i = 0; i < h; i++)scanf("%s", s[i]);
	puts(dfs(0, 0, h % 2) ? "YES" : "NO");

	return 0;
}