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:
n,m=map(int,raw_input().split())
ar=[]
for i in xrange(n):
	inp=map(int,raw_input().split())
	ar.append(inp)
ans=0
if False:
	for i in xrange(n):
		for j in xrange(m):
			if ar[i][j]==1:
				for k in xrange(0,n):
					if ar[k][j]==0:
						ans=1
						break
				for k in xrange(0,m):
					if ar[i][k]==0:
						ans=1
						break
			if ans==1:
				break
		if ans==1:
			break
mat=[]
for i in xrange(n):
	temp=[]
	for j in xrange(m):
		temp.append(1)
	mat.append(temp)
if True:
	for i in xrange(n):
		for j in xrange(m):
			if ar[i][j]==0:
				for k in xrange(0,n):
					mat[k][j]=0
						
				for k in xrange(0,m):
					mat[i][k]=0
	
ne=[]
for i in xrange(n):
	temp=[]
	for j in xrange(m):
		temp.append(0)
	ne.append(temp)
for i in xrange(n):
	for j in xrange(m):
		if mat[i][j]==1:
			for k in xrange(0,n):
				ne[k][j]=1
			for k in xrange(0,m):
				ne[i][k]=1

ans=0
for i in xrange(n):
	for j in xrange(m):
		if ar[i][j]!=ne[i][j]:
			ans=1
			break
	if ans==1:
		break
if ans==1:
	print "NO"
else:
	print "YES"
	for i in xrange(n):
		for j in xrange(m):
			print mat[i][j],
		print