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,cp[30],vis[30];
char s[30][30],t[30][30];
inline void copy(int x,int y){for(int i=0;i<m;i++)t[y][i]=s[x][i];}
bool dfs(int u)
{
	if(u==n)
	{
		for(int i=0,j=0;i<n;i++)
			if(cp[i]==n)copy(i,n/2);
			else if(cp[i]>i){copy(i,j),copy(cp[i],n-j-1);j++;}
		memset(vis,0,sizeof vis);
		for(int i=0,cnt=0;i<m;i++)
			if(!vis[i])
			{
				for(int j=i+1;j<m;j++)
					if(!vis[j])
					{
						bool f=1;
						for(int k=0;k<n;k++)
							if(t[k][i]!=t[n-k-1][j]){f=0;break;}
						if(f){vis[i]=vis[j]=1;break;}
					}
				if(!vis[i])
				{
					if(++cnt>m%2)return 0;
					for(int j=0;j<n;j++)
						if(t[j][i]!=t[n-j-1][i])return 0;
				}
			}
		return 1;
	}
	if(cp[u]!=-1)return dfs(u+1);
	for(int i=u+1;i<n;i++)
		if(cp[i]==-1)
		{
			cp[i]=u;cp[u]=i;
			if(dfs(u+1))return 1;
			cp[i]=cp[u]=-1;
		}
	return 0;
}
int main()
{
	cin>>n>>m;
	for(int i=0;i<n;i++)cin>>s[i];
	memset(cp,-1,sizeof cp);
	if(n&1)
	{
		for(int i=0;i<n;i++)
		{
			cp[i]=n;
			if(dfs(0)){puts("YES");return 0;}
			cp[i]=-1;
		}
		puts("NO");
	}
	else puts(dfs(0)?"YES":"NO");
	return 0;
}
