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:
m, n = list(map(int, input().split()))
b = [list(map(int, input().split())) for _ in range(m)]
b_t = list(zip(*b))

elem = set()
rows = set()
cols = set()
possible = True
row = 0
while row < m and possible:
    col = 0
    while col < n:
        element = b[row][col]
        elem.add(element)
        if element == 1:
            if not all(b[row][i] == 1 for i in range(n)) and not all(b_t[col][i] == 1 for i in range(m)):
                possible = False
                break 
            if all(b[row][i] == 1 for i in range(n)):
                rows.add(row)
            if all(b_t[col][i] == 1 for i in range(m)):
                cols.add(col)
        col += 1
    row += 1

if not possible:
    print("NO")
elif (len(rows) == 0 or len(cols) == 0) and len(elem) == 2:
    print("NO")
else:
    print("YES")
    out = []
    for row in range(m):
        curr_row = []
        for col in range(n):
            if col in cols and row in rows:
                curr_row.append('1')
            else:
                curr_row.append('0')
        out.append(curr_row)
    for row in out:
        print(" ".join(row))