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:
read = lambda: map(int, input().split())
n, m = read()
a = [list(read()) for i in range(n)]
pi = [1] * m; pj = [1] * n
ai = []; aj = []
flag = 0
for i in range(n):
    if a[i] == pi: ai.append(i)
    else:
        for j in range(m):
            if a[i][j] and [a[k][j] for k in range(n)] != pj:
                print('NO')
                exit()
            elif a[i][j]: aj.append(j)
f1 = ai == [] or aj == []
f2 = not (ai == [] and aj == [])
f3 = a == [pi for i in range(n)]
if f1 and f2 and not f3:
    print('NO')
    exit()
print('YES')
A = [[0] * m for i in range(n)]
if not f3:
    for i in ai:
        for j in aj:
            A[i][j] = 1
else: A = a
[print(*i) for i in A]