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 PYTHON3):

Solution:
r, c = map(int, input().split())

matrix = [0] * r

row_zero, col_zero = set(), set()
for i in range(r):
    row = map(int, input().split())
    matrix[i] = list(row)

for i, row in enumerate(matrix):
    for j, elem in enumerate(row):
        if elem == 0:
            row_zero.add(i)
            col_zero.add(j)

mid_matrix = [[1] * c for i in range(r)]
row_ones, col_ones = set(), set()

for i, row in enumerate(mid_matrix):
    for j, elem in enumerate(row):
        if i in row_zero:
            mid_matrix[i][j] = 0
        if j in col_zero:
            mid_matrix[i][j] = 0
        if mid_matrix[i][j] == 1:
            row_ones.add(i)
            col_ones.add(j)

final_matrix = [[None] * c for i in range(r)]
for i, row in enumerate(final_matrix):
    for j, elem in enumerate(row):
        if i in row_ones:
            final_matrix[i][j] = 1
        if j in col_ones:
            final_matrix[i][j] = 1
        if not final_matrix[i][j]:
            final_matrix[i][j] = 0

if matrix == final_matrix:
    print("YES")
    for i, row in enumerate(mid_matrix):
        row = map(str, row)
        print(' '.join(row))
else:
    print("NO")