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 PYTHON):

Solution:
m, n = map(int, raw_input().split())

matrix = []
for i in range(m):
	matrix.append(map(int, raw_input().split()))


def fill(answer, i, j, m, n):
	# fill in 0's at ith row and jth column. 
	# if you come across any ones report failure.
	# m --> number of rows
	# n --> number of columns
	for x in range(m):
		if answer[x][j] == 1:
			return False
		else:
			answer[x][j] = 0
	for x in range(n):
		if answer[i][x] == 1:
			return False
		else:
			answer[i][x] = 0
	return True

def seek(answer, i, j, m, n):
	# find a cell that is currently set to None 
	# in the ith row and jth column.
	for x in range(m):
		if answer[x][j] == 1:
			return True
	for x in range(n):
		if answer[i][x] == 1:
			return True

	for x in range(m):
		if answer[x][j] == None:
			answer[x][j] = 1
			return True
	for x in range(n):
		if answer[i][x] == None:
			answer[i][x] = 1
			return True
	return False

def process(matrix, m, n):
	answer = [[None for x in range(len(matrix[0]))] 
	for x in range(len(matrix))]

	# fill in the mandatory zeros.
	for i in range(len(matrix)):
		for j in range(len(matrix[0])):
			if matrix[i][j] == 0:
				res = fill(answer, i, j, m, n)
				if not res:
					return ("NO", None)

	for i in range(len(matrix)):
		for j in range(len(matrix[0])):
			if matrix[i][j] == 1:
				res = seek(answer, i, j, m, n)
				if not res:
					return ("NO", None)
	for i in range(len(matrix)):
		for j in range(len(matrix[0])):
			if answer[i][j] == None:
				answer[i][j] = 0
	return ("YES", answer)

	

res = process(matrix, m, n)
if res[0] == "NO":
	print "NO"
else:
	print "YES"
	for row in res[1]:
		print " ".join(map(str, row))