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.util.*;
import java.io.*;
public class Main
{
 
  public static void main(String args[])throws 
UnsupportedEncodingException, IOException, Exception
  {
   Reader.init(System.in);
   PrintWriter pw=new PrintWriter(System.out);
   int n=Reader.nextInt();
   int m=Reader.nextInt();
   char t[][]=new char[n][m];
   for(int i=0;i<n;i++)
   {
     t[i]=Reader.next().toCharArray();
   }
   boolean fl=false;
   for(int i=0;i<n;i++)
   {
     for(int j=0;j<m;j++)
     {
       if(t[i][j]=='S')
       {
         if(i-1>-1){if(t[i-1][j]=='W')fl=true;}
         if(i+1<n){if(t[i+1][j]=='W')fl=true;}
         if(j+1<m){if(t[i][j+1]=='W')fl=true;}
         if(j-1>-1){if(t[i][j-1]=='W')fl=true;}
         if(fl)break;
       }
     }
     if(fl)break;
   }
   if(fl)pw.println("No");
   else
   {
     pw.println("Yes");
     for(int i=0;i<n;i++)
     {
       for(int j=0;j<m;j++)
       {
         if(t[i][j]!='S'&&t[i][j]!='W')
         pw.print("D");
         else pw.print(t[i][j]);
       }
       pw.println();
     }
   }
  pw.close();
}
  private static long gcd(long a, long b) {
    while (b > 0) {
      long temp = b;
      b = a % b;
      a = temp;
    }
    return a;
  }
 
  private static long lcm(long a, long b) {
    return a * (b / gcd(a, b));
  }
  
  
  
  public static boolean isPrime(long n)
  {
    
        boolean flag = false;
        for(int i = 2; i <= n/2; ++i)
        {
            if(n% i == 0)
            {
                flag = true;
                break;
            }
        }

        if (!flag)
            return true;
        else
            return false;
  }

}
class Pair<S extends Comparable<S>, T extends Comparable<T>> implements Comparable<Pair<S, T>> {
    S first;
    T second;
    
    Pair(S f, T s) {
        first = f;
        second = s;
    }

    @Override
    public int compareTo(Pair<S, T> o) {
        int t = first.compareTo(o.first);
        if (t == 0) return second.compareTo(o.second);
        return t;
    }
    
    @Override
    public int hashCode() {
        return (31 + first.hashCode()) * 31 + second.hashCode();
    }
    
    @Override
    public boolean equals(Object o) {
        if (!(o instanceof Pair)) return false;
        if (o == this) return true;
        Pair p = (Pair) o;
        return first.equals(p.first) && second.equals(p.second);
    }
    
    @Override
    public String toString() {
        return "Pair{" + first + ", " + second + "}";
    }
}
class Reader {
 
    static BufferedReader reader;
    static StringTokenizer tokenizer;
 
    static void init(InputStream input) throws UnsupportedEncodingException {
        reader = new BufferedReader(
                new InputStreamReader(input, "UTF-8"));
        tokenizer = new StringTokenizer("");
    }
 
    static void init(String url) throws FileNotFoundException {
        reader = new BufferedReader(new FileReader(url));
        tokenizer = new StringTokenizer("");
    }
 
    static String next() throws IOException {
        while (!tokenizer.hasMoreTokens()) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                    reader.readLine());
        }
        return tokenizer.nextToken();
    }
 
    static String nextLine() throws IOException {
        return reader.readLine();
    }
 
    static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
 
    static double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
 
    static long nextLong() throws IOException {
        return Long.parseLong(next());
    }
}