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:
def zip_sorted(a,b):

	# sorted  by a
	a,b = zip(*sorted(zip(a,b)))
	# sorted by b
	sorted(zip(a, b), key=lambda x: x[1])

	return a,b

def column(matrix, i):
    return [row[i] for row in matrix]

m,n = [int(n1) for n1 in input().split()]
b   = []
for i in range(m):
	a   = [int(n1) for n1 in input().split()]
	b.append(a)

c = [[1 for x in range(n)] for y in range(m)]

d = [1]*m
e = [1]*n

d1 = [0]*m
e1 = [0]*n

count = 0

for i in range(len(b)):

	for j in range(len(b[i])):

		if b[i][j]==0:

			d[i] = 0
			e[j] = 0


if ((count>=1) and ((sum(d)+sum(e))==0)):

	print('NO')

else:

	for i in range(len(c)):

		for j in range(len(c[i])):

			if d[i]==1 and e[j]==1:
				
				c[i][j] = 1

			else:

				c[i][j] = 0


	count = 0

	for i in range(len(c)):

		for j in range(len(c[i])):

			if (b[i][j]==1):

				if ((1 in c[i]) or (1 in column(c, j))):

					pass

				else:

					count = 1

					break

		if count==1:

			break

	if count==0:

		if sum(sum(c,[]))==0 and count>=1 :

			print('NO')

		else:

			print('YES')

			for i in range(len(c)):

				print(*c[i])
	else:

		print('NO')
'''
for i in range(n):
for j in range(n):
for k1 in range(len(a)):
for k2 in range(len(a)):
for k3 in range(len(a)):

'''