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[13],b[13];
string s[13],ss[13];
void dfs1(int x,bool allow){
	if(x==m){cout<<"YES"<<endl;exit(0);}
	else if(b[x]!=-1) dfs1(x+1,allow);
	else{
		for(int i=x+1;i<m;i++){
			if(b[i]!=-1) continue;
			bool ok=true;
			for(int j=0;j<n;j++)
				if(s[j][x]!=s[a[j]][i]){ok=false;break;}
			if(ok){
				b[x]=i;b[i]=x;dfs1(x+1,allow);b[x]=-1;b[i]=-1;
			}
		}
		if(allow){b[x]=x;dfs1(x+1,0);b[x]=-1;}
	}
} 
void dfs2(int x,bool allow){
	if(x==n) dfs1(0,(m%2));
	else if(a[x]!=-1) dfs2(x+1,allow);
	else{
		for(int i=x+1;i<n;i++)
			if(ss[i]==ss[x]&&a[i]==-1){
				a[x]=i;a[i]=x;dfs2(x+1,allow);a[x]=-1;a[i]=-1;
			}
		if(allow){a[x]=x;dfs2(x+1,0);a[x]=-1;}
	}
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>m;
	for(int i=0;i<n;i++) cin>>s[i];
	memset(a,-1,sizeof(a));memset(b,-1,sizeof(b));
	for(int i=0;i<n;i++)
		ss[i]=s[i],sort(ss[i].begin(),ss[i].end());
	dfs2(0,(n%2));
	cout<<"NO"<<endl;
	return 0;
}
