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 arr_inp():
    return [int(x) for x in stdin.readline().split()]


def fill_mem():
    for i in range(m):
        for j in range(n):
            if mat[i][j] == 0:
                row[i], col[j] = 1, 1


def valid():
    r, c = 0, 0
    for i in row.values():
        if i:
            r += 1
    for i in col.values():
        if i:
            c += 1

    for i in range(m):
        for j in range(n):
            if mat[i][j]:
                if (r == m or c == n):
                    exit(print('NO'))
                if col[j] and row[i]:
                    exit(print('NO'))


def print_ans():
    print('YES')
    for i in range(m):
        for j in range(n):
            if not row[i] and not col[j]:
                print(1, end=' ')
            else:
                print(0, end=' ')
        print(end='\n')


from sys import stdin
from collections import defaultdict

m, n = arr_inp()
mat, row, col = [arr_inp() for i in range(m)], defaultdict(int), defaultdict(int)
fill_mem()
valid()
print_ans()
# print(row, col)