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>
using namespace std;
int n,m,a[15],b[15];
char g[15][15];
bool check(int x)
{
	if(x>=m) return 1;
	if(~b[x]) return check(x+1);
	int tmp=0;
	for(int i=x;i<m;i++) tmp^=!(~b[i]);
	for(int i=x;i<m;i++) 
	{
		if(!(~b[i]))
		{
			b[i]=x;
			b[x]=i;
			bool flg=1;
			for(int j=0;j<n;j++) if(g[j][x]!=g[a[j]][i]) flg=0;
			if(i==x&&flg&&tmp&&check(x+1)) return 1;
			if(i!=x&&flg) return check(x+1);
			b[x]=b[i]=-1;
		}
	}
	return 0;
}
bool dfs(int x)
{
	if(x>=n) return check(0);
	if(~a[x]) return dfs(x+1);
	int tmp=0;
	for(int i=x;i<n;i++) tmp^=!(~a[i]);
	for(int i=x;i<n;i++)
	{
		if(!(~a[i]))
		{
			a[i]=x;
			a[x]=i;
			if((tmp||i!=x)&&dfs(x+1)) return 1;
			a[i]=a[x]=-1;
		}
	}
	return 0;
}
int main()
{
	cin>>n>>m;
	memset(a,-1,sizeof(a));
	memset(b,-1,sizeof(b));
	for(int i=0;i<n;i++) for(int j=0;j<m;j++) cin>>g[i][j];
	if(dfs(0)) puts("YES");
	else puts("NO");
	return 0;
}