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<vector>
#include<algorithm>
#include<string>

using namespace std;

int h, w;
vector<string> s;
vector<int> hp;
bool ans = false;

bool isMatch(int a, int b){
    for(int i = 0; i < h; i++){
        if(s[i][a] != s[hp[i]][b]) return false;
    }
    return true;
}

void check(){
    int wOdd = w % 2;
    vector<int> used(w, 0);
    for(int i = 0; i < w; i++){
        if(used[i]) continue;
        for(int j = i + 1; j < w; j++){
            if(used[j]) continue;
            if(isMatch(i, j)){
                used[i] = 1; used[j] = 1;
                break;
            }
        }
    }
    if(count(used.begin(), used.end(), 0) > 1) return;
    if(wOdd){
        for(int i = 0; i < w; i++){
            if(used[i]) continue;
            if(isMatch(i, i)){
                break;
            }else{
                return;
            }
        }
    }
    ans = true;
}

void dfs(int y, bool hOdd){
    if(ans) return;
    if(y >= h){
        check();
        return;
    }
    if(hp[y] != -1){
        dfs(y + 1, hOdd);
        return;
    }
    if(hOdd){
        hp[y] = y;
        dfs(y + 1, false);
        hp[y] = -1;
    }
    for(int i = y + 1; i < h; i++){
        if(hp[i] != -1) continue;
        hp[y] = i; hp[i] = y;
        dfs(y + 1, hOdd);
        hp[y] = -1; hp[i] = -1; 
    }
    hp[y] = -1;
}

int main(){
    
    cin >> h >> w;
    s.resize(h);
    for(int i = 0; i < h; i++){
        cin >> s[i];
    }
    hp = vector<int>(h, -1);
    if(h % 2){
        dfs(0, true);
    }else{
        dfs(0, false);
    }
    if(ans){
        cout << "YES" << endl;
    }else{
        cout << "NO" << endl;
    }
    return 0;
}