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:
m,n = map(int,raw_input().split())

    
matrix_B = {}
#matrix_A = {key:[-1]*n for key in xrange(m)}
matrix_A = {key:[ 0 ] * n for key in xrange( m )}
rows = []
cols = [0] * n
flag = False
wrong_A = False

for i in xrange( m ):
    matrix_B[ i ] = map(int,raw_input().split())
    rows.append( sum( matrix_B[ i ] ) )

for row in xrange( m ):
    for col in xrange( n ):
        cols[ col ] += matrix_B[ row ][ col ]

zero_count = 0
for r in xrange( m ):
    
    for c in xrange( n ):
    
        if matrix_B[ r ][ c ] == 1 and rows[ r ] != n and cols[ c ] != m:
            wrong_A  = True
            break
                    
        if rows[ r ] == n and cols[ c ] == m:
                       
            matrix_A[ r ][ c ] = 1
            flag = True
        elif rows[ r ] == 0 and cols[ c ] == 0:
            zero_count += 1
    if wrong_A:
        flag = False
        break
        
if flag or zero_count == m * n:
    print "YES"
    for row in matrix_A.values():
        for elt in row:
            print elt,
        print ""
else:
    print "NO"
       

##    if sum( matrix_B[ row ] ) == n:
##        
##        for cl in xrange( n ):
##            sm = 0
##            for rw in xrange( m )
##                sm += matrix_B[ rw ][ cl ]
##
##            if sm == m:
##                matrix_A[    



##for i in xrange(m):
##    for j in xrange(n):
##        if matrix_B[i][j] == 0:
##            for row in xrange(m):
##                if matrix_A[row][j] != -1 and matrix_A[row][j] != 1:
##                    matrix_A[row][j] = 0
##
##            for col in xrange(n):
##                if matrix_A[i][col] != -1 and matrix_A[row][j] != 1:
##                    matrix_A[i][col] = 0
##            
##        else:
##            for row in xrange(m):
##                if matrix_A[row][j] != -1:
##                    matrix_A[row][j] = 1
##
##            for col in xrange(n):
##                if matrix_A[i][col] != -1:
##                    matrix_A[i][col] = 1
##
##for key in matrix_A:
##    print " ".join(matrix_A[key])
##            
##