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 = []
for i in range(0,m):
	row = [int(x) for x in input().split(' ')]
	B.append(row)

B_ = []
for j in range(0,n):
	col = [B[i][j] for i in range(0,m)]
	B_.append(col)

for row in B:
	if 1 in row:
		if 0 in row:
			for j in range(0,n):
				col = [B[i][j] for i in range(0,m)]
				if (row[j]==1 and 0 in col):
					print("NO")
					exit()

for col in B_:
	if 1 in col:
		if 0 in col:
			for i in range(0,m):
				row = [B[i][j] for j in range(0,n)]
				if (col[i]==1 and 0 in row):
					print("NO")
					exit()

only1 = False
only0 = False
for row in B:
	if 1 not in row:
		only0 = True 
	if 0 not in row:
		only1 = True
if only1 and only0:
	print("NO")
	exit()

only1 = False
only0 = False
for col in B_:
	if 1 not in col:
		only0 = True 
	if 0 not in col:
		only1 = True
if only1 and only0:
	print("NO")
	exit()

A = [[0 for col in range(0,n)] for row in range(0,m)]
R = []
for i in range(0,m):
	flg = 1
	for j in range(0,n):
		if B[i][j] == 0:
			flg = 0
			break
	if flg == 1:
		R.append(i)
C = []
for j in range(0,n):
	flg = 1
	for i in range(0,m):
		if B[i][j] == 0:
			flg = 0
			break
	if flg == 1:
		C.append(j)

for r in R:
	for c in C:
		A[r][c] = 1

print("YES")
for i in range(0,m):
	s = ""+str(A[i][0])
	for j in range(1,n):
		s += " "+str(A[i][j])
	print(s)