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:
n, m = map(int, raw_input().split())
L = [[1] * m for _ in xrange(n)]
B = []
for _ in xrange(n):
    B.append(map(int, raw_input().split()))

for i in xrange(n):
    for j in xrange(m):
        if B[i][j] == 0:
            for k in xrange(n):
                L[k][j] = 0
            for k in xrange(m):
                L[i][k] = 0

row = set()
column = set()

for i in xrange(n):
    for j in xrange(m):
        if L[i][j] == 1:
            row.add(i)
            column.add(j)
no_flag = False
for i in xrange(n):
    for j in xrange(m):
        if B[i][j] == 1:
            if i not in row and j not in column:
                no_flag = True
                break
    if no_flag:
        break

if no_flag:
    print "NO"
else:
    print "YES"
    for i in xrange(n):
        print ' '.join(map(str, L[i]))