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:
def print_values(values,m,n):
    for i in range(m):
        for j in range(n):
            print values[i][j],
        print " "
        
def compute_col_output(values,m,n):
    out = 1;
    for i in range(m):
        out = out and values[i][n]
    return out

def compute_col_or(values,m,n):
    out = 0
    for i in range(m):
        out = out or values[i][n]
    return out

def calculate_or_matrix(output,calculated,m,n):
    for i in range(m):
        for j in range(n):
            row_calculated = any(output[i])
            col_calculated = compute_col_or(output,m,j)
            if row_calculated or col_calculated:
                calculated[i][j] = 1

m,n = map(int, raw_input().split())
values = []
output = []
calculated = []
for i in range(m):
    values.append(map(int, raw_input().split()))
    output.append([0]*n)
    calculated.append([0]*n)


for i in range(m):
    for j in range(n):
        if values[i][j] ==  1:
            row_output = all(values[i])
            col_output = compute_col_output(values,m,j)
            if row_output and col_output:
                output[i][j] = 1
            else:
                output[i][j] = 0

calculate_or_matrix(output,calculated,m,n)
if calculated == values:
    print "YES"
    print_values(output,m,n)
else:
    print "NO"