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]=[int(x) for x in input().split()]

matrix=[]

for _ in range(n):
	l=[int(x) for x in input().split()]
	matrix.append(l)

def operation_or(l):

	row_op=[]
	col_op=[]
	output=[]
	for i in range(n):
		temp=1
		for j in range(m-1):
			temp=temp and matrix[i][j] and matrix[i][j+1]
		row_op.append(temp)

	for j in range(m):
		temp=1
		for i in range(n-1):
			temp=temp and matrix[i][j] and matrix[i+1][j]
		col_op.append(temp)		
	for i in range(n):
		temp=[]
		for j in range(m):
			temp.append(row_op[i] and col_op[j])
		output.append(temp)
	return output	

def operation_and(l):

	row_op=[]
	col_op=[]
	output=[]
	for i in range(n):
		temp=0
		for j in range(m-1):
			temp=temp or l[i][j] or l[i][j+1]
			# print(temp)
		row_op.append(temp)
	# print(row_op)	
	for j in range(m):
		temp=0
		for i in range(n-1):
			temp=temp or l[i][j] or l[i+1][j]
		col_op.append(temp)		
	# print(col_op)
	for i in range(n):
		temp=[]
		for j in range(m):
			temp.append(row_op[i] or col_op[j])
		output.append(temp)
	return output	
if n!=1 or m!=1:

	output=operation_or(matrix)
	# print(output)
	op_cmp=operation_and(output)
	# print(op_cmp)
	flag=False

	for i,j in zip(op_cmp,matrix):
		for q,k in zip(i,j):
			if q!=k:
				flag=True
				break
	if flag==True: print('NO')			
	if flag!=True:
		print('YES')
		for i in output:
			for j in i:
				print(j,end=" ")
			print()	
else:
	print('YES')
	print(matrix[0][0])