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 <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
using namespace std;
const int N = 15;
char s[N][N];bool vis[N];
int mark[N],match[N],n,m,dt;
bool solve()
{
	bool mid = m & 1;
	for(int i = 1;i <= m;i++)match[i] = 0;
	for(int i = 1;i <= m;i++)if(!match[i])
	{
		for(int j = i + 1;j <= m && !match[i];j++)if(!match[j])
		{
			bool f = true;
			for(int k = 1;k <= n && f;k++)if(s[mark[k]][i] != s[mark[n - k + 1]][j])f = false;
			if(f)match[i] = j,match[j] = i;
		}
		if(!match[i] && mid)
		{
		  bool f = true;
		  for(int k = 1;k <= n && f;k++)if(s[mark[k]][i] != s[mark[n - k + 1]][i])f = false;
		  if(f)match[i] = i,mid = false;
		}
		if(!match[i])return false;
	}
	return true;
}
bool dfs(int now,int pos,bool mid)
{
	if(now > n)return solve();
	if(vis[now])return dfs(now + 1,pos,mid);
	for(int i = now + 1;i <= n;i++)if(!vis[i])
	{
		mark[pos] = i;mark[n - pos + 1] = now;
		vis[i] = vis[now] = true;
		if(dfs(now + 1,pos + 1,mid))return true;
		mark[pos] = mark[n - pos +1] = 0;
		vis[i] = vis[now] = false;
	}
	if(mid){mark[n >> 1 | 1] = now;return dfs(now + 1,pos,0);}
	return false;
}
int main()
{
  scanf("%d %d",&n,&m);
  for(int i = 1;i <= n;i++)scanf("%s",s[i] + 1);
  if(dfs(1,1,n & 1))puts("YES");else puts("NO");
}