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:
import itertools
n,m = map( int , raw_input().split())
mm = []
for row in range(n):
    split = map( int, raw_input().split())
    mm.append(split)
    
def row_sum(n,m,i):
    sum = 0
    for ii in range(m): sum+= mm[i][ii]
    return sum    

def col_sum(n,m,i):
    sum = 0
    for ii in range(n): sum+= mm[ii][i]
    return sum

filled_rows = []
filled_columns = []
for xx in range(n):
    for yy in range(m):
        
        val = mm[xx][yy]

        if val == 1:
            c_sum = col_sum(n,m,yy)
            r_sum = row_sum(n,m,xx)

            if not (c_sum == n or r_sum == m): 
                print "NO" 
                exit(0)

            if c_sum == n: 
                if yy not in filled_columns: filled_columns.append(yy)
            if r_sum == m: 
                if xx not in filled_rows: filled_rows.append(xx)

if (len(filled_columns) == 0  or len(filled_rows) == 0) and len(filled_columns) + len(filled_rows) > 0:     
    print "NO"
    exit(0)

print "YES"
mprime = []
ones = list(itertools.product(filled_rows,filled_columns))
for row in range(n):
    for col in range(m):
        if (row,col) in ones: print 1, 
        else: print 0,
    print " "