Problem: Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus decided to place some shepherd dogs in such a way that all his sheep are protected.

The pasture is a rectangle consisting of R × C cells. Each cell is either empty, contains a sheep, a wolf or a dog. Sheep and dogs always stay in place, but wolves can roam freely around the pasture, by repeatedly moving to the left, right, up or down to a neighboring cell. When a wolf enters a cell with a sheep, it consumes it. However, no wolf can enter a cell with a dog.

Initially there are no dogs. Place dogs onto the pasture in such a way that no wolf can reach any sheep, or determine that it is impossible. Note that since you have many dogs, you do not need to minimize their number. 

Input

First line contains two integers R (1 ≤ R ≤ 500) and C (1 ≤ C ≤ 500), denoting the number of rows and the numbers of columns respectively.

Each of the following R lines is a string consisting of exactly C characters, representing one row of the pasture. Here, 'S' means a sheep, 'W' a wolf and '.' an empty cell.

Output

If it is impossible to protect all sheep, output a single line with the word "No".

Otherwise, output a line with the word "Yes". Then print R lines, representing the pasture after placing dogs. Again, 'S' means a sheep, 'W' a wolf, 'D' is a dog and '.' an empty space. You are not allowed to move, remove or add a sheep or a wolf.

If there are multiple solutions, you may print any of them. You don't have to minimize the number of dogs.

Examples

Input

6 6
..S...
..S.W.
.S....
..W...
...W..
......


Output

Yes
..SD..
..SDW.
.SD...
.DW...
DD.W..
......


Input

1 2
SW


Output

No


Input

5 5
.S...
...S.
S....
...S.
.S...


Output

Yes
.S...
...S.
S.D..
...S.
.S...

Note

In the first example, we can split the pasture into two halves, one containing wolves and one containing sheep. Note that the sheep at (2,1) is safe, as wolves cannot move diagonally.

In the second example, there are no empty spots to put dogs that would guard the lone sheep.

In the third example, there are no wolves, so the task is very easy. We put a dog in the center to observe the peacefulness of the meadow, but the solution would be correct even without him.

Solution (in JAVA):

Solution:
import java.awt.Point;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.*;
import java.util.*;

public class Main {
	//////
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	//////
	public static long modularExponentiation(long x, long n, long M) {
		long result = 1;
		while (n > 0) {
			if (n % 2 == 1)
				result = (result * x) % M;
			x = (x * x) % M;
			n = n / 2;
		}
		return result;
	}

	public static long modularExponentiation1(long x, long n) {
		long result = 1;
		while (n > 0) {
			if (n % 2 == 1)
				result = (result * x);
			x = (x * x);
			n = n / 2;
		}
		return result;
	}

	/////
	
	static public int fun(int num){
		int prod=1;
		while(num>0){
			if(num%10!=0){
			prod=prod*(num%10);
			}
			num=num/10;
		}
		return prod;
	}
	
	public static boolean dfs(int x,int find,boolean[] visited,int[][] matrix){
		
		boolean ans=false; 
		if(x==find){
			ans=true;
			return true;
		}
		for(int i=0;i<26;i++){
			if(visited[i]==false&&matrix[x][i]==1){
				visited[i]=true;
				ans=ans||dfs(i,find,visited,matrix);
			}
		}
		
		return ans;
	}
	
	static int lis(int arr[],int n,int k,int b)
    {
          int lis[] = new int[n];
          int i,j,max = 0;
 
          /* Initialize LIS values for all indexes */
           for ( i = 0; i < n; i++ )
              lis[i] = 1;
 
           /* Compute optimized LIS values in bottom up manner */
           for ( i = 1; i < n; i++ )
              for ( j = 0; j < i; j++ ) 
                         if ( arr[i] >= k*arr[j]+b && lis[i] < lis[j] + 1)
                    lis[i] = lis[j] + 1;
 
           /* Pick maximum of all LIS values */
           for ( i = 0; i < n; i++ )
              if ( max < lis[i] )
                 max = lis[i];
 
            return max;
    }
	
	public static void main(String[] args) throws IOException {

		FastReader scan = new FastReader();
		PrintWriter pw = new PrintWriter(System.out);
		
		int N=scan.nextInt();
		int M=scan.nextInt();
		
		char[][] dp=new char[N][M];
		for(int i=0;i<N;i++){
				String str=scan.next();
				dp[i]=str.toCharArray();
		}
		int flag=0;
		for(int i=0;i<N;i++){
			for(int j=0;j<M;j++){
				if(dp[i][j]=='S'){
					if(j-1>=0){
						if(dp[i][j-1]=='.'){
							dp[i][j-1]='D';
						}
					}
					if(j+1<M){
						if(dp[i][j+1]=='.'){
							dp[i][j+1]='D';
						}
					}
					if(i-1>=0){
						if(dp[i-1][j]=='.'){
							dp[i-1][j]='D';
						}
					}
					if(i+1<N){
						if(dp[i+1][j]=='.'){
							dp[i+1][j]='D';
						}
					}
					if(j-1>=0){
						if(dp[i][j-1]=='W'){
							flag=1;
						}
					}
					if(j+1<M){
						if(dp[i][j+1]=='W'){
							flag=1;
						}
					}
					if(i-1>=0){
						if(dp[i-1][j]=='W'){
							flag=1;
						}
					}
					if(i+1<N){
						if(dp[i+1][j]=='W'){
							flag=1;
						}
					}

					
					

				}
			}
		}
		
		
		if(flag==1){
			System.out.println("No");
		}else{
			System.out.println("Yes");
			for(int i=0;i<N;i++){
				for(int j=0;j<M;j++){
					System.out.print(dp[i][j]);
				}
				System.out.println();
			}
		}
		
		
		
	}
}