Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

2 3
arc
rac

Sample Output 1

YES

If the second and third columns from the left are swapped, the grid becomes symmetric, as shown in the image below:

Sample Input 2

3 7
atcoder
regular
contest

Sample Output 2

NO

Sample Input 3

12 12
bimonigaloaf
faurwlkbleht
dexwimqxzxbb
lxdgyoifcxid
ydxiliocfdgx
nfoabgilamoi
ibxbdqmzxxwe
pqirylfrcrnf
wtehfkllbura
yfrnpflcrirq
wvcclwgiubrk
lkbrwgwuiccv

Sample Output 3

YES

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<algorithm>
#include<bitset>
#include<cctype>
#include<cstring>
#include<cstdlib>
#include<deque>
#include<fstream>
#include<functional>
#include<iomanip>
#include<iostream>
#include<map>
#include<math.h>
#include<queue>
#include<set>
#include<stack>
#include<stdio.h>
#include<sstream>
#include<string>
#include<time.h>
#include<utility>
#include<vector>

using namespace std;

int n,m;
bool ans;
string s[15],t[15];

void make(){
	map<string,int> cnt;
	for(int i=0;i<m;i++){
		string str;
		for(int j=0;j<n;j++){
			if(t[i][j]=='*') str+=t[n-i-1][j];
			else str+=t[i][j];
		}
		string trs=str;
		reverse(trs.begin(),trs.end());
		cnt[min(trs,str)]++;
	}
	bool flag=false;
	for(map<string,int>::iterator it=cnt.begin();it!=cnt.end();it++){
		if(!(it->second%2)) continue;
		string str=it->first;
		string rts=str;
		reverse(rts.begin(),rts.end());
		if(str!=rts){
			if(flag) return;
			flag=true;
		}
	}
	ans=true;
}

void dfs(int b,int k){
	if(!b){
		make();
		return;
	}
	int x;
	while(~b>>x&1) x++;
	t[k]=s[x];
	for(int i=x+1;i<n;i++){
		if(!(b&(1<<i))) continue;
		t[n-k-1]=s[i];
		dfs(b^(1<<x)^(1<<i),k+1);
	}
}

int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++) cin>>s[i];
	if(n&1) s[n]=string(m,'*');
	dfs((1<<n)-1,0);
	if(ans) puts("YES");
	else puts("NO");
	return 0;
}