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:
MAX = 201
global B
global A
global n,m
B = [[0 for i in xrange(MAX)] for i in xrange(MAX)]
A = [[0 for i in xrange(MAX)] for i in xrange(MAX)]

def linhaEColuna(i,j):

	for k in xrange(n):
		if B[k][j] == 0: return False
	for k in xrange(m):
		if B[i][k] == 0: return False
	
	return True

def linhaOuColuna(i,j):
	
	aux1 = aux2 = 1
	for k in xrange(n):
		if B[k][j] == 0: aux1 = 0
	for k in xrange(m):
		if B[i][k] == 0: aux2 = 0

	return aux1 + aux2;

n,m = map(int,raw_input().split())
for i in xrange(n):
	B[i] = map(int,raw_input().split())

flag = True
flag2 = True
cont = 0
for i in xrange(n):
	for j in xrange(m):
		if B[i][j] == 1 and linhaEColuna(i,j):
			A[i][j] = 1
			cont += 1
		if B[i][j] == 1 and linhaOuColuna(i,j) == 0:
			flag = False
		if B[i][j] == 1:
			flag2 = False
if flag and (cont > 0 or flag2):
	print "YES"
	for i in xrange(n):
		for j in xrange(m):
			print A[i][j],
		print
else:
	print "NO"