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 math, sys
from collections import defaultdict, Counter, deque
from heapq import heapify, heappush, heappop
from bisect import bisect_left, bisect_right

INF = float('inf')
MOD = int(1e9) + 7
MAX = int(2e6)

def check(mat, n, m):
	ans = [[0 for i in range(m)] for j in range(n)]

	rows = [0 for i in range(n)]
	cols = [0 for i in range(m)]

	for i in range(n):
		for j in range(m):
			if mat[i][j] == 1:
				rows[i] = 1
				cols[j] = 1

	for i in range(n):
		for j in range(m):
			ans[i][j] = rows[i] | cols[j]

	return ans 

def solve():
	n, m = vars()

	mat = []

	for i in range(n):
		mat.append(array())

	rows = [1 for i in range(n)]
	cols = [1 for i in range(m)]

	for i in range(n):
		for j in range(m):
			if mat[i][j] == 0:
				rows[i] = 0

	for j in range(m):
		for i in range(n):
			if mat[i][j] == 0:
				cols[j] = 0

	ans = [[0 for i in range(m)] for j in range(n)]
	for i in range(n):
		for j in range(m):
			ans[i][j] = rows[i] & cols[j]

	if mat == check(ans, n, m):
		print('YES')
		for i in range(n):
			for j in range(m):
				print(ans[i][j], end=" ")
			print()
	else:
		print('NO')
	
def main():
	t = 1 
	# t = int(input())
	for _ in range(t):
		solve()
















	
def gcd(a, b):
	while b:
		a, b = b, a % b
	return a 

def sieve():
	primes = [ 1 for i in range(MAX) ]
	primes[0] = primes[1] = 0 
	i = 2
	while i <= MAX ** 0.5:
		j = i * i 
		while primes[i] and j < MAX:
			if j % i == 0:
				primes[j] = 0 
			j += i 
		i += 1
	return primes

def input():
	return sys.stdin.readline().rstrip('\n').strip()

def print(*args, sep=' ', end='\n'):
	first = True 
	for arg in args:
		if not first:
			sys.stdout.write(str(sep))
		sys.stdout.write(str(arg))
		first = False 

	sys.stdout.write(end)


def vars():
	return map(int, input().split())

def array():
	return list(map(int, input().split()))


if __name__ == '__main__':

	main()