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,n) for(int i = 0; i < n; i++)
int h, w, ar[15];
bool used[15], ev;
char s[12][12];
inline bool isRev(char a[], char b[]){
	rep(i,h) if(a[i] != b[h-1-i]) return false;
	return true;
}
bool dfs(int p, int n, int e){
	if(e){
		rep(i,h){
			used[i] = true; ar[h/2] = i;
			if(dfs(0,0,0)) return true;
			used[i] = false;
		}
		return false;
	}else if(n != h/2){
		while(used[p]) p++;
		for(int i = p+1; i < h; i++) if(!used[i]){
			used[i] = true;
			ar[n] = p; ar[h-1-n] = i;
			if(dfs(p+1, n+1, 0)) return true;
			used[i] = false;
		}
		return false;
	}else{
		char temp[15][15];
		bool u[15] = {}, ev = !(w%2);
		rep(i,w) rep(j,h) temp[i][j] = s[ar[j]][i];
		rep(i,w){
			for(int j = i+1; j < w && !u[i]; j++) if(!u[j])
				u[i] = u[j] = isRev(temp[i], temp[j]);
			if(!u[i] && !ev) u[i] = ev = isRev(temp[i], temp[i]);
			if(!u[i]) return false;
		}
		return true;
	}
}
int main(){
	scanf("%d%d", &h, &w);
	rep(i,h) scanf("%s", s[i]);
	puts(dfs(0,0,h%2) ? "YES" : "NO");
}