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 exit
m, n = map(int, input().split())
row = [True] * m
col = [True] * n
a = [[0] * n for _ in range(m)]
for i in range(m):
    s = list(map(int, input().split()))
    for j in range(n):
        if (s[j] == 0):
            row[i] = False
            col[j] = False
        a[i][j] = s[j]
c = 0
r = 0
for i in range(n):
    if col[i]:
        c+=1
for j in range(m):
    if row[j]:
        r+=1
res = []
if (r > 0 and c > 0) or (r == 0 and c == 0):
    for i in range(m):
        res.append([])
        for j in range(n):
            if col[j] and row[i]:
                res[i].append(1)
            elif a[i][j] == 0 or (col[j] or row[i]):
                res[i].append(0)
            else:
                print('NO')
                exit(0)

    print('YES')
    for i in range(m):
        s = ' '.join(list(map(str, res[i])))
        print(s)
else:
    print('NO')