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 PYTHON):

Solution:
"""
Author    : raj1307 - Raj Singh
Institute : Jalpaiguri Government Engineering College
Date      : 13.05.19
"""
from __future__ import division, print_function
import itertools,os,sys

from io import BytesIO, IOBase

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
else:
    from builtins import str as __str__
    str = lambda x=b'': x if type(x) is bytes else __str__(x).encode()

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._buffer = BytesIO()
        self._fd = file.fileno()
        self._writable = 'x' in file.mode or 'r' not in file.mode
        self.write = self._buffer.write if self._writable else None

    def read(self):
        return self._buffer.read() if self._buffer.tell() else os.read(self._fd, os.fstat(self._fd).st_size)

    def readline(self):
        while self.newlines == 0:
            b, ptr = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)), self._buffer.tell()
            self._buffer.seek(0, 2), self._buffer.write(b), self._buffer.seek(ptr)
            self.newlines += b.count(b'\n') + (not b)
        self.newlines -= 1
        return self._buffer.readline()

    def flush(self):
        if self._writable:
            os.write(self._fd, self._buffer.getvalue())
            self._buffer.truncate(0), self._buffer.seek(0)


sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip(b'\r\n')


def print(*args, **kwargs):
    sep, file = kwargs.pop('sep', b' '), kwargs.pop('file', sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop('end', b'\n'))
    if kwargs.pop('flush', False):
        file.flush()



def ii(): return int(input())
def si(): return str(input())
def mi(): return map(int,input().strip().split(" "))
def li(): return list(mi())

from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify, #heappop ,heappush, heapreplace
#from math import ceil,floor,log,sqrt,factorial,pow,pi
#from bisect import bisect_left,bisect_right
#from decimal import *,threading

def dmain():
    sys.setrecursionlimit(100000000)
    threading.stack_size(40960000)
    thread = threading.Thread(target=main)
    thread.start()

abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[0] 
def sort2(l):return sorted(l, key=getKey)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [int(i) for i in str(n)]
def powerMod(x,y,p):
    res = 1
    x %= p
    while y > 0:
        if y&1:
            res = (res*x)%p
        y = y>>1
        x = (x*x)%p1
    return res
def gcd(x, y):
    while y:
        x, y = y, x % y
    return x

# For getting input from input.txt file 
#sys.stdin = open('input.txt', 'r')  
  
# Printing the Output to output.txt file 
#sys.stdout = open('output.txt', 'w') 

def valid(x,y,n,m):
    if x<0 or x>=n or y<0 or y>=m or l[x][y]=='S':
        return 1
    elif l[x][y]=='W':
        return 2
    
    else:
        return 3



n,m=mi()
l=[]
for i in range(n):
    l.append(list(si()))

def main():
    
    for i in range(n):
        for j in range(m):
            if l[i][j]=='S':
                p=valid(i+1,j,n,m)
                if p==3:
                    l[i+1][j]='D'
                elif p==2:
                    print('No')
                    exit()
                
                p=valid(i-1,j,n,m)
                if p==3:
                    l[i-1][j]='D'
                elif p==2:
                    print('No')
                    exit()
                
                p=valid(i,j+1,n,m)
                if p==3:
                    l[i][j+1]='D'
                elif p==2:
                    print('No')
                    exit()
                
                p=valid(i,j-1,n,m)
                if p==3:
                    l[i][j-1]='D'
                elif p==2:
                    print('No')
                    exit()
    
    
    print('Yes')
    for i in range(n):
        print(''.join(l[i]))
    
    
    
    
    
    
    
    
    

if __name__ == "__main__":
    main()
    #dmain()