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:
#fin, fout = open("input.txt"), open("output.txt", "w")
n, m = map(int, input().split())
#n, m = map(int, fin.readline().split())
A = [list(map(int, input().split())) for i in range(n)]
#A = [list(map(int, fin.readline().split())) for i in range(n)]
I, J = [0] * n, [0] * m
IS, JS = [0] * n, [0] * m
CountI, CountJ = 0, 0
for i in range(n):
	IS[i] = sum(A[i])
	if (IS[i] == m):
		I[i] = 1
		CountI += 1
for j in range(m):
	JS[j] = sum([A[i][j] for i in range(n)])
	if (JS[j] == n):
		J[j] = 1
		CountJ += 1
flag = True
i = 0
while i < n and flag:
	if (IS[i] != m and IS[i] != CountJ):
		flag = False
	i += 1
j = 0
while j < m and flag:
	if (JS[j] != n and JS[j] != CountI):
		flag = False
	j += 1
if (flag and not (sum(I) + sum(J) > 0 and sum(I) * sum(J) == 0)):
	print("YES")
	for i in range(n):
		for j in range(m):
			print(I[i] * J[j], end = " ")
		print()
else:
	print("NO")
#fout.close()
#fin.close()