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:
from sys import stdin, stdout
from collections import Counter



def no_mistake(m, n, A, B):
    for i in range(m):
        for j in range(n):
            if B[i][j] == 1:
                flag = False
                if A[i].__contains__(1):
                    flag = True
                    continue
                else:
                    for k in range(m):
                        if A[k][j] == 1:
                            flag = True
                            break
                if not flag:
                    return False
    return True



def format_output_list(alist):
    res = ""
    for element in alist:
        res += str(element) + " "
    res = res.strip()
    res += "\n"
    return res


def format_matrix(m, n, A):
    res = ""
    for alist in A:
        res += format_output_list(alist)
    return res


def solve(m, n, B):
    A = [[1]*n]*m
    for i in range(m):
        for j in range(n):
            if B[i][j] == 0:
                A[i] = [0]*n
                for k in range(m):
                    A[k][j] = 0
    if no_mistake(m, n, A, B):
        return "YES\n" + format_matrix(m, n, A)
    return "NO"



if __name__ == "__main__":
    m, n = list(map(int, stdin.readline().rstrip().split()))
    count = m
    matrix = []
    while count > 0:
        count -= 1
        matrix.append(list(map(int, stdin.readline().rstrip().split())))
    stdout.write(str(solve(m, n, matrix)))