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:
import sys
def input(): return sys.stdin.readline().strip()

m, n = map(int, input().split())
b = [0 for i in range(m)]

for i in range(m):
	b[i] = [int(i) for i in input().split()]

row = [0 for i in range(m)]
col = [0 for i in range(n)]

for i in range(m):
	for j in range(n):
		if not b[i][j]:
			row[i] = 1
			col[j] = 1


a = [[1 for i in range(n)]for j in range(m)]
zero = [0 for i in range(n)]
row_val = [0 for i in range(m)]
col_val = [0 for i in range(n)]

for i in range(m):
	if row[i]:
		a[i] = zero

for i in range(n):
	if col[i]:
		for j in range(m):
			a[j][i] = 0

for i in range(m):
	for j in range(n):
		if a[i][j]:
			col_val[j] = 1
			row_val[i] = 1

possible = True
for i in range(m):
	for j in range(n):
		if b[i][j] != (row_val[i] or col_val[j]):
			possible = False
			break

if possible:
	print("YES")
	for i in range(m):
		for j in range(n):
			print(a[i][j], end=" ")
		print()
else:
	print("NO")