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 <stdio.h>
#include <string.h>
#include <algorithm>
#define rep(i,st,ed) for (register int i=st;i<=ed;++i)

const int N=25;

char s[N][N];

int match[N],n,m,flag;

bool used[N],lxf;

bool check() {
	int cnt=m;
	rep(i,1,m) used[i]=0;
	rep(i,1,m) if (!used[i]) {
		rep(j,i+1,m) if (!used[j]) {
			bool wjp=true;
			rep(k,1,n) if (s[k][i]!=s[match[k]][j]) {
				wjp=false; break;
			}
			if (wjp) {
				used[i]=used[j]=true;
				cnt-=2;
				break;
			}
		}
	}
	if (cnt==0) return true;
	if (cnt==1&&m%2==1) {
		rep(i,1,m) if (!used[i]) {
			rep(j,1,n) if (s[j][i]!=s[match[j]][i]) return false;
			break;
		}
		return true;
	}
	return false;
}

void dfs(int dep) {
	if (flag) return ;
	if (dep==(n+1)/2+1) return (void) (flag|=check());
	rep(i,1,n) if (!match[i]) {
		rep(j,i+1,n) if (!match[j]) {
			match[i]=j; match[j]=i;
			dfs(dep+1);
			match[i]=match[j]=0;
		}
		if ((n&1)&&!lxf) {
			match[i]=i; lxf=true;
			dfs(dep+1);
			match[i]=0; lxf=false;
		}
	}
}

int main(void) {
	scanf("%d%d",&n,&m);
	rep(i,1,n) scanf("%s",s[i]+1);
	dfs(1);
	puts(flag?"YES":"NO");
	return 0;
}