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:
'''
Created on 16/11/2014

@author: DAVID
'''

import sys

A = []
B = []

line = [int(i) for i in sys.stdin.readline().split()]
m = line[0]
n = line[1]

for i in range( m ):
    line = [int(j) for j in sys.stdin.readline().split()]
    B.append( line )
    tmp = []
    for j in range( len( line ) ):
        tmp.append( 1 )
    A.append( tmp )
    
for i in range( m ):
    for j in range( n ):
        if B[i][j] == 0:
            for k in range( m ):
                A[k][j] = 0
            for k in range( n ):
                A[i][k] = 0
       

for i in range( m ):
    for j in range( n ):
        b = False
        if B[i][j] == 1:
            for k in range( m ):
                if A[k][j] == 1:
                    b = True
                    break;
            for k in range( n ):
                if A[i][k] == 1:
                    b = True
                    break
        else:
            b = True
        if not b:
            break
    if not b:
        break
    
if b:
    print "YES"
    for i in range( m ):
        for j in range( n ):
            print A[i][j],
        print 
else:
    print "NO"