Hongcow likes solving puzzles.

One day, Hongcow finds two identical puzzle pieces, with the instructions "make a rectangle" next to them. The pieces can be described by an n by m grid of characters, where the character 'X' denotes a part of the puzzle and '.' denotes an empty part of the grid. It is guaranteed that the puzzle pieces are one 4-connected piece. See the input format and samples for the exact details on how a jigsaw piece will be specified.

The puzzle pieces are very heavy, so Hongcow cannot rotate or flip the puzzle pieces. However, he is allowed to move them in any directions. The puzzle pieces also cannot overlap.

You are given as input the description of one of the pieces. Determine if it is possible to make a rectangle from two identical copies of the given input. The rectangle should be solid, i.e. there should be no empty holes inside it or on its border. Keep in mind that Hongcow is not allowed to flip or rotate pieces and they cannot overlap, i.e. no two 'X' from different pieces can share the same position.

Input

The first line of input will contain two integers n and m (1 ≤ n, m ≤ 500), the dimensions of the puzzle piece.

The next n lines will describe the jigsaw piece. Each line will have length m and will consist of characters '.' and 'X' only. 'X' corresponds to a part of the puzzle piece, '.' is an empty space.

It is guaranteed there is at least one 'X' character in the input and that the 'X' characters form a 4-connected region.

Output

Output "YES" if it is possible for Hongcow to make a rectangle. Output "NO" otherwise.

Examples

Input

2 3
XXX
XXX


Output

YES


Input

2 2
.X
XX


Output

NO


Input

5 5
.....
..X..
.....
.....
.....


Output

YES

Note

For the first sample, one example of a rectangle we can form is as follows 
    
    
      
    111222  
    111222  
    

For the second sample, it is impossible to put two of those pieces without rotating or flipping to form a rectangle.

In the third sample, we can shift the first tile by one to the right, and then compose the following rectangle: 
    
    
      
    .....  
    ..XX.  
    .....  
    .....  
    .....  
    


import java.io.BufferedReader;

import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;


import java.util.*;

public class random_num {
public static long x,y,d;
public static int mod= (int) (Math.pow(10,9)+7);
	public static void main(String args[]){
		InputReader s= new InputReader(System.in);
		OutputStream outputStream= System.out;
		PrintWriter out= new PrintWriter(outputStream);
		int n=s.nextInt();
		int m=s.nextInt();
		TreeSet<Integer> x=new TreeSet();
		TreeSet<Integer> y=new TreeSet();
		int a[][]=new int[n][m];
		for(int i=0;i<n;i++){
			String str=s.nextLine();
			for(int j=0;j<m;j++){
				a[i][j]=str.charAt(j);
			}
		}
		int f=0;
		int mini=0,minj=0,flag=0;
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
				if(a[i][j]=='X'){
					x.add(i);
					y.add(j);
				}
			}
		}
		//System.out.println(x.first()+" "+x.last());
		//System.out.println(y.first()+" "+y.last());
		for(int i=x.first();i<=x.last();i++){
			for(int j=y.first();j<=y.last();j++){
				if(a[i][j]!='X'){
					f=1;
					break;
				}
			}
			if(f==1){
				break;
			}
		}
		if(f==0)
			System.out.println("YES");
		else
			System.out.println("NO");
		out.close();
	}
	static class InputReader {
		
	    public BufferedReader reader;
	    public StringTokenizer tokenizer;
 
	    public InputReader(InputStream inputstream) {
	      reader = new BufferedReader(new InputStreamReader(inputstream));
	      tokenizer = null;
	    }
	    
	    public String nextLine(){
	    	String fullLine=null;
	    	while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	              fullLine=reader.readLine();
	            } catch (IOException e) {
	              throw new RuntimeException(e);
	            }
	            return fullLine;
	          }
	          return fullLine;
	    }
		public String next() {
	      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	        try {
	          tokenizer = new StringTokenizer(reader.readLine());
	        } catch (IOException e) {
	          throw new RuntimeException(e);
	        }
	      }
	      return tokenizer.nextToken();
	    }
		public long nextLong() {
		      return Long.parseLong(next());
		    }
	    public int nextInt() {
	      return Integer.parseInt(next());
	    }
	  }
	
	public static HashMap<Integer,Long> getcatalan(int n){
		HashMap<Integer,Long> a=new HashMap<Integer,Long>();
		a.put(0,(long)1);
		for(int i=1;i<=n+1;i++){
			long m=2*(2*i-1);
			long m1=m*a.get(i-1);
			long m3=m1/(i+1);
			//out.println(m+" "+m1+" "+m2+" "+m3);
			
			a.put(i,m3%1000000007);
		}
		return a;
	}
	
	public static int getGdc(int[] x) {
	    // get the smallest of all number no need to check for higher values
	    int smallest = getSmallest(x);

	    for(int i = smallest; i >= 1; i--) {
	       int j;
	       for(j = 0; j < x.length; ++j) {
	           if(x[j] % i != 0)
	               break;
	       }
	       // if we pass through the array with all % == 0 return the value
	       if(j == x.length)
	           return i;
	    }
	    // so the only possible is 1
	    return 1;
	}

	// return smallest number of an array of int
	public static int getSmallest(int[] x) {
	    int smallest = x[0];
	    for(int i = 1; i < x.length; ++i) {
	        if(x[i] < smallest)
	            smallest = x[i];
	    }
	    return smallest;
	}
	public static long gcd(long number1, long number2) {  
		if(number2 == 0){
			return number1;
		} 
		return gcd(number2, number1%number2); 
	}
	
	public static void extendedEuclidean(long a,long b){
		if(b == 0) {
	        d = a;
	        x = 1;
	        y = 0;
	    }
	    else {
	        extendedEuclidean(b, a%b);
	        long temp = x;
	        x = y;
	        y = temp - (a/b)*y;
	    }
	}
	
	public static int combinations(int n,int r){
		if(n==r) return 1;
		if(r==1) return n;
		if(r==0) return 1;
		return combinations(n-1,r)+ combinations(n-1,r-1);
	}
	public static long binomialCoeff(int n, int k)
	{
	    long C[][]= new long[n+1][k+1];
	    int i, j;
	 
	    // Caculate value of Binomial Coefficient in bottom up manner
	    for (i = 0; i <= n; i++)
	    {
	        for (j = 0; j <= Math.min(i, k); j++)
	        {
	            // Base Cases
	            if (j == 0 || j == i)
	                C[i][j] = 1;
	 
	            // Calculate value using previosly stored values
	            else
	                C[i][j] = C[i-1][j-1] + C[i-1][j];
	        }
	    }
	 
	    return C[n][k];
	}
	public static long expo(long a, long b){
	    if (b==1)
	        return a%mod;
	    if (b==2)
	        return (a%mod)*(a%mod);

	    if (b%2==0){
	            return expo(expo(a%mod,(b%mod)/2),2);
	    }
	    else{
	        return a*(expo(expo(a%mod,((b%mod-1%mod)%mod)/2),2));
	    }
	}
	
	public static int[] sieve(int N){
		int arr[]= new int[N+1];
		int b[]=new int[N+1];
		for(int i=2;i<Math.sqrt(N);i++){
			if(arr[i]==0){
				
				for(int j= i*i;j<= N;j= j+i){
					arr[j]=1;
				}
			}
		}
		return arr;
		// All the i for which arr[i]==0 are prime numbers.
	}
	static class Pair implements Comparable<Pair>
    {
        int f,s;
        Pair(int ii, int cc)
        {
            f=ii;
            s=cc;
        }
        
        public int compareTo(Pair o) 
        {
            return Integer.compare(this.f, o.f);
        }
        
    }
	public static class company{
		public int t;
		public int d;
		public company(int m,int f){
			this.t=m;
			this.d=f;
		}
		public int get_t(){
			return this.t;
		}
		public int get_d(){
			return this.d;
		}
	}
	public static class pair_comparator implements Comparator<company>{

		@Override
		public int compare(company arg0, company arg1) {
			// TODO Auto-generated method stub
			return arg0.t-arg1.t;
		}
		
	}

}