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:
m,n=map(int,input().split())
B=[list(map(int,input().split()))for i in range(m)]

sr=[sum(B[i])for i in range(m)]
sc=[sum([B[i][j]for i in range(m)]) for j in range(n)]

no=sum(sr)>0 and(max(sr)<n or max(sc)<m)

for i in range(m):
 for j in range(n):
  if B[i][j] and sc[j]%m and sr[i]%n or no:    #The main part is writing NO condition
   print('NO')
   exit()
print('YES')
for i in range(m):
  Ai=[(sr[i]//n)*(sc[j]//m)for j in range(n)]   #print 1 if corresponding rows qnd columns are full of 1s
  print(' '.join(repr(a)for a in Ai))