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:
"""three conditions:

 1> if all row has a -->  0
 	or all col has a --> 0
 	and if we have 1 any where --->  not possible

 2> if a element is 1
    and zero in row as well as col ---->not posssible

 3> yes

  	if col and row  has no zero then it's ===> 1
  	else ===> 0

       """

r,c = map(int,input().split())

a = [list(map(int,input().split())) for i in range(r)]

ans = []

# lets calculate sum of rows an sum of col



sum_row = [sum(a[i]) for i in range(r)]
sum_col = [sum(a[i][j] for i in range(r)) for j in range(c)]


#condition 1
no = sum(sum_row) > 0 and  (max(sum_row)<c or max(sum_col)<r )

	


#condition 2
for i in range(r):
	for j in range(c):

		if a[i][j] and sum_row[i]<c and sum_col[j] < r or no:
			
			print("NO")
			exit()
print("YES")
ans = [[(sum_row[i]//c)*(sum_col[j]//r) for j in range(c)] for i in range(r)]
for i in range(r):
	print(*ans[i])