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 = []
original = [[0 for _ in range (c)] for _ in range (r)]
idex = 0
rowCheck = [0 for _ in range(r)]
columCheck = [0 for _ in range(c)]
ans = ['YES','NO']
for _ in range(r):
    matrix.append(list(map(int,input().split())))
put = 0
row = []
for i in range(r):
    if 0 not in matrix[i]:
        rowCheck[i] = 1
for j in range(c):
    start = 1
    for i in range(r):
        start&=matrix[i][j]
    if start == 1:
        columCheck[j] = 1
for i in range(r):
    if rowCheck[i]:
        for j in range(c):
            if columCheck[j]:
                original[i][j] = 1
rowCheck2 = [0 for _ in range(r)]
columCheck2 = [0 for _ in range(c)]
for i in range(r):
    if 1 in original[i]:
        rowCheck2[i] = 1
for j in range(c):
    start = 0
    for i in range(r):
        start|=original[i][j]
    if start == 1:
        columCheck2[j] = 1
for i in range(r):
    if idex:
        break
    for j in range(c):
        if matrix[i][j]!= rowCheck2[i]|columCheck2[j]:
            idex = 1
if idex:
    print(ans[idex])
else:
    print(ans[idex])
    for v in original:
        print(*v)