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

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

def checkFirstRow(matrix):
	cols = []

	for i in range(len(matrix[0])):
		if matrix[0][i] == 1:
			validity = checkCol(matrix, i)
			if not validity:
				for k in range(len(matrix[0])):
					if matrix[0][k] == 0:
						print "NO"
						exit()
			else:
				cols.append(i)

	return cols

def checkFirstCol(matrix):
	rows = []

	for i in range(len(matrix)):
		if matrix[i][0] == 1:
			validity = checkRow(matrix, i)
			if not validity:
				for k in range(len(matrix)):
					if matrix[k][0] == 0:	
						print "NO"
						exit()
			else:
				rows.append(i)
	return rows

def checkEquality(matrix1, matrix2):
	for i in range(len(matrix1)):
		for j in range(len(matrix1[i])):
			if matrix1[i][j] != matrix2[i][j]:
				print "NO"
				exit()

def finalChecker(matrixA, matrixB):
	matrixC = [[0 for k in range(numCols)] for j in range(numRows)]
	for i in range(len(matrixA)):
		for j in range(len(matrixA[i])):
			if (matrixA[i][j] == 1):
				for m in range(len(matrixC[i])):
					matrixC[i][m] = 1
				for n in range(len(matrixC)):
					matrixC[n][j] = 1
	checkEquality(matrixB, matrixC)

def makeResult(rows, cols):
	matrixA = [[0 for k in range(numCols)] for j in range(numRows)]
	
	for e in rows:
		for e2 in cols:
			matrixA[e][e2] = 1

	return matrixA
	

numRows, numCols = raw_input().split()
numRows = int(numRows)
numCols = int(numCols)

b = []

for i in range(numRows):
	b.append(map(int, raw_input().split()))

cols = checkFirstRow(b)
rows = checkFirstCol(b)

if (len(cols) == 0 and len(rows) > 0) or (len(rows) == 0 and len(cols) > 0):
	print "NO"
	exit()

matrixA = makeResult(rows,cols)
finalChecker(matrixA, b)

print "YES"

for row in matrixA:
	output = ""
	for col in range(len(row) - 1):
		output += str(row[col]) + " "
	output += str(row[len(row) - 1])
	print output