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:
import sys
input = sys.stdin.readline
read_tuple = lambda _type: map(_type, input().split(' '))
from math import factorial
    
def cross_fill(i, j, m, n, matrix, value):
    for idx in range(m):
        matrix[idx][j] = value
    for idx in range(n):
        matrix[i][idx] = value

def _or(i, j, m, n, matrix):
    res = 0
    for idx in range(m):
        res += matrix[idx][j]
    for idx in range(n):
        res += matrix[i][idx]
    return int(res > 0)


def create_B(matrix, m, n):
    res = [[None for _ in range(n)] for _ in range(m)]
    for i in range(m):
        for j in range(n):
            res[i][j] = _or(i, j, m, n, matrix)
    return res

def solve():
    m, n = read_tuple(int)
    B = [list(read_tuple(int)) for _ in range(m)]
    A = [[1 for _ in range(n)] for _ in range(m)]
    for i in range(m):
        for j in range(n):
            if B[i][j] == 0:
                cross_fill(i, j, m, n, A, 0)
    reproduced_B = create_B(A, m, n)
    if B == reproduced_B:
        print("YES")
        for row in A:
            print(' '.join(map(str, row)))
    else:
        print("NO")

    
if __name__ == '__main__':
    solve()