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 SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;
typedef long double ld;
int n, m;
int a[15], used[15], res;
char s[15][15];
void f(int cur) {
	if(cur == n + 1) {
		bool usedMiddle = 0;
		for(int i=1; i<=m; i++) used[i] = 0;
		for(int i=1; i<=m; i++) {
			if(used[i]) continue;
			bool found = 0;
			for(int j=i+1; j<=m; j++) {
				if(used[j]) continue;
				bool flag = 1;
				for(int k=1; k<=n; k++)
					if(s[k][i] != s[a[k]][j] || s[a[k]][i] != s[k][j])
						{ flag = 0; break; }
				if(flag) {
					used[i] = used[j] = 1;
					found = 1;
					break;
				}
			}
			if(!found) {
				if(m % 2 == 0) return;
				if(usedMiddle) return;
				bool flag = 1;
				for(int k=1; k<=n; k++)
					if(s[k][i] != s[a[k]][i])
						{ flag = 0; break; }
				if(!flag) return;
				used[i] = 1;
				usedMiddle = 1;
			}
		}
		res = 1;
		return;
	}
	if(a[cur] != 0) {
		f(cur + 1);
		return;
	}
	for(int i=cur+1; i<=n; i++) {
		if(a[i] != 0) continue;
		a[cur] = i;
		a[i] = cur;
		f(cur + 1);
		a[cur] = a[i] = 0;
	}
}
int main() {
	scanf("%d%d", &n, &m);
	for(int i=1; i<=n; i++)
		scanf("%s", &s[i][1]);
	if(n & 1) {
		for(int i=1; i<=n; i++) {
			a[i] = i;
			f(1);
			a[i] = 0;
		}
	} else f(1);
	puts(res ? "YES" : "NO");
	return 0;
}