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:
string = input()
numbers = string.split()
a, b =  int(numbers[0]), int(numbers[1])
ma = [[1 for x in range(b)] for y in range(a)]
mb = []
for x in range(a):
    string = input()
    mb.append(list(map(int, string.split())))
for x in range(a):
    for y in range(b):
        if mb[x][y] == 0:
            for z in range(b):
                ma[x][z] = 0
            for z in range(a):
                ma[z][y] = 0
results = "YES"
for x in range(a):
    for y in range(b):
        if mb[x][y] == 1:
            for z in range(b):
                if ma[x][z]:
                    break
            else:
                results = "NO"
            if results == "NO":
                for z in range(a):
                    if ma[z][y]:
                        results = "YES"
                        break
                else:
                    results = "NO"
                    break
    if results == "NO":
        break
print(results)
if results == "YES":
    for x in ma:
        print(" ".join(map(str, x)))