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
//ΔARC 095C
#include<iostream>
#include<cstdio>
#include<fstream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;
typedef double DB;
const int N = 15;
int n,m,p[N];
char a[N][N];
string l[N],r[N];
int main()
{
	srand(19260817);
	int i,j,x,f,o=30000;
	cin>>n>>m;
	for(i=1;i<=n;i=i+1)
		scanf("%s",a[i]+1);
	while(o--){
		for(j=1;j<=m;j=j+1)
			l[j]="",r[j]="";
		for(i=1;i<=n;i=i+1)
			p[i]=i;
		random_shuffle(p+1,p+n+1);
		for(i=1;i<=n;i=i+1){
			x=p[i];
			for(j=1;j<=m;j=j+1)
				l[j]=l[j]+a[x][j],r[j]=a[x][j]+r[j];
		}
		for(j=1;j<=m;j=j+1)
			p[j]=0;
		f=m&1;
		for(j=1;j<=m;j=j+1){
			if(p[j])
				continue;
			p[j]=1;
			x=1;
			while(p[x]||x<=m&&l[j]!=r[x]){
				x++;
			}
			if(x<=m)
				p[x]=1;
			else{
				if(l[j]==r[j]&&f)
					f--;
				else
					break;
			}
		}
		if(j>m){
			cout<<"YES";
			return 0;
		}
	}
	cout<<"NO";
	return 0;
}