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;

#define rep(i,l,r) for(int i=l;i<=r;++i)
const int N=12+1;
int n,m;
char s[N][N];
int dy[N];

bool same(int c1,int c2)
{
	rep(r,0,n-1)
	if(s[r][c1]!=s[dy[r]][c2])return 0;
	return 1;
}
void check()
{
	bool can=m%2;
	bool have[N]={};
	rep(c,0,m-1)
	if(!have[c])
	{
		rep(c2,c+1,m-1)
		if(same(c,c2))
		{	
			have[c]=have[c2]=1;
			break;
		}
		if(!have[c])
		{
			if(!can||!same(c,c))return ;
			have[c]=1;
			can=0;
		}
	}
	puts("YES");
	exit(0);
}

void dfs(int x,bool can)
{
	if(x>=n){check();return ;}
	if(dy[x]!=-1){dfs(x+1,can);return ;}
	if(can){dy[x]=x;dfs(x+1,0);}
	int y=x;
	while(++y<n)
	if(dy[y]==-1)
	{
		dy[x]=y;
		dy[y]=x;
		dfs(x+1,can);
		dy[y]=-1;
	}
	dy[x]=-1;
}

int main()
{
	cin>>n>>m;
	rep(i,0,n-1)scanf("%s",s[i]);
	rep(i,0,n-1)dy[i]=-1;
	dfs(0,n%2);
	puts("NO");
}