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:
n, m = map(int,input().split())
b = []
for i in range(n):
	b.append( list(map(int,input().split())) )
a = [[1 for i in range(m)] for i in range(n)]

row, col = set(),set()
for i in range(n):
	for j in range(m):
		if b[i][j] == 0:
			row.add(i)
			col.add(j)
			

for r in row:
	for i in range(m):
		a[r][i] = 0
for c in col:
	for i in range(n):
		a[i][c] = 0

# a is the required the matrix 
# now try to regenerate b with this a, if possible then YES else NO
gFlag = True
for i in range(n):
	for j in range(m):
		# if its a 1 -> then either col has 1 or row has 1 in a
		if b[i][j] == 1:
			valid = False
			# row
			for k in range(m):
				if a[i][k]==1:
					valid = True
					break
			# col
			for k in range(n):
				if a[k][j] == 1:
					valid = True
					break
			if not valid:
				gFlag = False
				break
		else:
			# ith row and jth col must have all zeros
			valid = True
			# row
			for k in range(m):
				if a[i][k]==1:
					valid = False
					break
			# col
			for k in range(n):
				if a[k][j]==1:
					valid = False
					break
			if not valid:
				gFlag = False
				break
	if not gFlag:
		break
if not gFlag:
	print("NO")
else:
	print("YES")
	for i in range(n):
		for j in range(m):
			print(a[i][j],end=" ")
		print()