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, k = [int(x) for x in input().split()]
arr = []
ans = []
rows = [0] * n
cols = [0] * k
for i in range(n):
	arr.append(input().split())
	ans.append(['0'] * k)
	rows[i] = all(map(lambda a: a == '1', arr[-1]))
for i in range(k):
	sum = 0
	for j in range(n):
		sum += 1 if arr[j][i] == '1' else 0
	if sum == n:
		cols[i] = True
for i in range(n):
	for j in range(k):
		if ((arr[i][j] == '1' and not rows[i] and not cols[j]) or
                   (arr[i][j] == '0' and (rows[i] or cols[j]))):
			print('NO')
			exit()
if any(rows) ^ any(cols):
	print('NO')
	exit()
for i in range(n):
	if not rows[i]:
		continue
	for j in range(k):
		if cols[j]:
			ans[i][j] = '1'
print('YES')
print('\n'.join(' '.join(l) for l in ans))