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:
import fileinput

# utils

def read_int_array(string):
	return map(lambda x: int(x), string.strip().split(' '))

# input
stdin = fileinput.input()
source = []
for line in stdin:
	source.append(line)
# source = open('data.txt', 'r').readlines()

# solution 

m, n = read_int_array(source[0])
source = source[1:]
b = []
a = []
for line in source:
	b.append(read_int_array(line))
	a.append([0]*n)

def is_row_of_ones(row, matrix):
	for i in matrix[row]:
		if i != 1:
			return False
	return True

def is_col_of_ones(col, matrix):
	for row in range(len(matrix)):
		if matrix[row][col] != 1:
			return False
	return True

def eval_b(a):
	b = []
	for row in a:
		b.append([0] * len(row))

	for i in range(m):
		for j in range(n):
			if a[i][j] == 1:
				for k in range(m):
					b[k][j] = 1
				for k in range(n):
					b[i][k] = 1
	return b


for i in range(m):
	for j in range(n):
		if is_row_of_ones(i, b) and is_col_of_ones(j, b):
			a[i][j] = 1

if b == eval_b(a):
	print 'YES'
	for i in range(m):
		print ' '.join(map(lambda x: str(x), a[i]))

else:
	print 'NO'