Problem: Let's define logical OR as an operation on two logical values (i. e. values that belong to the set {0, 1}) that is equal to 1 if either or both of the logical values is set to 1, otherwise it is 0. We can define logical OR of three or more logical values in the same manner:

<image> where <image> is equal to 1 if some ai = 1, otherwise it is equal to 0.

Nam has a matrix A consisting of m rows and n columns. The rows are numbered from 1 to m, columns are numbered from 1 to n. Element at row i (1 ≤ i ≤ m) and column j (1 ≤ j ≤ n) is denoted as Aij. All elements of A are either 0 or 1. From matrix A, Nam creates another matrix B of the same size using formula:

<image>.

(Bij is OR of all elements in row i and column j of matrix A)

Nam gives you matrix B and challenges you to guess matrix A. Although Nam is smart, he could probably make a mistake while calculating matrix B, since size of A can be large.

Input

The first line contains two integer m and n (1 ≤ m, n ≤ 100), number of rows and number of columns of matrices respectively.

The next m lines each contain n integers separated by spaces describing rows of matrix B (each element of B is either 0 or 1).

Output

In the first line, print "NO" if Nam has made a mistake when calculating B, otherwise print "YES". If the first line is "YES", then also print m rows consisting of n integers representing matrix A that can produce given matrix B. If there are several solutions print any one.

Examples

Input

2 2
1 0
0 0


Output

NO


Input

2 3
1 1 1
1 1 1


Output

YES
1 1 1
1 1 1


Input

2 3
0 1 0
1 1 1


Output

YES
0 0 0
0 1 0

Solution (in PYTHON3):

Solution:
r,c=map(int,input().split())
b=[]
rcnt=0
ccnt=0
for i in range(r):
  x=list(map(int,input().split()))
  b.append(x)
a=[[0 for i in range(c)] for j in range(r)]
cmp={}
rmp={}
for i in range(r):
  rmp[i]=0
  count=0
  for j in range(c):
    if(b[i][j]==1):
      count+=1
    else:
      break    
  if(count==c):
    rmp[i]=1
    rcnt+=1
        
for i in range(c):
  cmp[i]=0
  count=0
  for j in range(r):
    if(b[j][i]==1):
      count+=1  
    else:
      break  
  if(count==r):
    cmp[i]=1  
    ccnt+=1 

chk=0
for i in range(r):
  for j in range(c):
    if(b[i][j]==1):
      if(rmp[i]==1 or cmp[j]==1):
        continue
      else:
        chk=1
        break

if(chk!=1):
  if(rcnt==ccnt or(rcnt>=1 and ccnt>=1)):        
    print("YES")
    for i in rmp.keys():
      for j in cmp.keys():
        if(rmp[i]==1 and cmp[j]==1):
          a[i][j]=1
    for i in range(r):
      for j in range(c):
        print(a[i][j],end=" ")
      print()
  else:
    print("NO")    
else:
  print("NO")