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: Accepted)

#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,cnt1,cnt2;
char s[15][15],b[15][15],c[15][15];
bool used[15],use[15];
int row[15][30],col[15][30];

int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++){
		scanf("%s",s[i]);
	}
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			row[i][s[i][j]-'a']++;
		}
	}
	for(int i=0;i<m;i++){
		for(int j=0;j<n;j++){
			col[i][s[j][i]-'a']++;
		}
	}
	int diff1=0,diff2=0;
	for(int i=0;i<n;i++){
		if(used[i]) continue;
		bool Ok=0;
		for(int j=0;j<n;j++){
			if(used[j]||j==i) continue;
			bool ok=1;
			for(int k=0;k<26;k++){
				if(row[i][k]!=row[j][k]){
					ok=0;
					break;
				}
			}
			if(ok){
				used[i]=1;
				used[j]=1;
				for(int k=0;k<m;k++){
					b[cnt1][k]=s[i][k];
					b[n-1-cnt1][k]=s[j][k];
				}
				cnt1++;
				Ok=1;
				break;
			}
		}
		if(!Ok){
			diff1++;
		}
	}
	if(n%2&&diff1==1){
		for(int i=0;i<n;i++){
			if(used[i]) continue;
			for(int j=0;j<m;j++){
				b[n/2][j]=s[i][j];
			}
		}
	}
//	for(int i=0;i<n;i++){
//		printf("%s\n",b[i]);
//	}
	if(n%2&&diff1>1){
		puts("NO");
		return 0;
	}else if((!(n%2))&&diff1>0){
		puts("NO");
		return 0;
	}
	for(int i=0;i<m;i++){
		if(use[i]) continue;
		bool Ok=0;
		for(int j=0;j<m;j++){
			if(use[j]||j==i) continue;
			bool ok=1;
			for(int k=0;k<26;k++){
				if(col[i][k]!=col[j][k]){
					ok=0;
					break;
				}
			}
			if(ok){
				for(int k=0;k<n;k++){
//					printf("%d %d %d %d\n",k,cnt2,k,i);
//					printf("%d %d %d %d\n",k,m-1-cnt2,k,j);
					c[k][cnt2]=b[k][i];
					c[k][m-1-cnt2]=b[k][j];
				}
				cnt2++;
				use[i]=1;
				use[j]=1;
				Ok=1;
				break;
			}
		}
		if(!Ok){
			diff2++;
		}
	}
	if(m%2&&diff2==1){
		for(int i=0;i<m;i++){
			if(use[i]) continue;
			for(int j=0;j<n;j++){
				c[j][m/2]=b[j][i];
			}
		}
	}
	if(m%2&&diff2>1){
		puts("NO");
		return 0;
	}else if((!(m%2))&&diff2>0){
		puts("NO");
		return 0;
	}
	bool ok=1;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			if(c[i][j]!=c[n-1-i][m-1-j]){
				ok=0;break;
			}
		}
	}
//	for(int i=0;i<n;i++){
//		for(int j=0;j<m;j++) printf("%c",c[i][j]);
//		puts("");
//	}
	if(!ok){
		puts("NO");
		return 0;
	}
//	printf("%d %d\n",diff1,diff2);
	puts("YES");
	return 0;
}