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:
while True:
	try:
		def isvalid(r, c, mt, tmt):
			a = b = d = e = 0
			valid = True
			while a < r:
				while b < c:
					if mt[a][b] == 1:
						flag = False
						while d < r:
							if tmt[d][b] == 1:
								flag = True
							d += 1
						d = 0
						while e < c:
							if tmt[a][e] == 1:
								flag = True
							e += 1
						e = 0
						if not flag:
							valid = False
					b += 1
				a += 1
				b = 0
				if not valid:
					break
			
			if not valid:print("NO")
			else:
				print("YES")
				for i in range(r):
					print(*tmt[i])
				
								
		def solution(r, c, mt):
			
			tmt = [[1 for i in range(c)]for j in range(r)]
			a = 0;b = 0;cc = 0;d = 0
			while a < r:
				while b < c:
					if mt[a][b] == 0:
						while cc < c:
							tmt[a][cc] = 0
							cc += 1
						cc = 0
						while d < r:
							tmt[d][b] = 0
							d += 1
						d = 0
						
					b += 1
				a += 1
				b = 0
				
			isvalid(r, c, mt, tmt)	
			
	
		def read():
			m, n = map(int, input().split())
			mtrx = [list(map(int, input().split()))for i in range(m)]
			
			solution(m, n, mtrx)
			
		if __name__ == "__main__":
			read()
	except EOFError:
		break