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:

def main():

    # m - строки
    # n - столбцы
    m, n = [int(i) for i in input().split()]

    b = [[int(i) for i in input().split()] for i in range(m)]
    a = [[1 for _ in range(n)] for _ in range(m)]

    for bi in range(m):
        for bj in range(n):
            if b[bi][bj] == 1:
                continue

            # change col values
            for j in range(n):
                a[bi][j] = 0

            # change row values
            for i in range(m):
                a[i][bj] = 0

    valid = True
    for ai in range(m):
        for aj in range(n):

            status = False # there is one
            for j in range(n):
                if a[ai][j] == 1:
                    status = True

            for i in range(m):
                if a[i][aj] == 1:
                    status = True

            if status:
                if b[ai][aj] != 1:
                    valid = False
            else:
                if b[ai][aj] != 0:
                    valid = False

    if valid:
        print('YES')

        for row in a:
            print(*row)

    else:
        print('NO')


main()