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:
r,c = map(int,input().split())
grid = [list(map(int,input().split())) for _ in range(r)]

r_count = [0 for _ in range(r)]
c_count = [0 for _ in range(c)]

for i,r_ in enumerate(grid):
    r_count[i] = r_.count(1)
    for j,c_ in enumerate(r_):
        if c_ == 1: c_count[j] +=1

out = [[0 for _ in range(c)] for row in range(r)]
entry = False
for x in range(r):
    for y in range(c):
        if r_count[x] == c and c_count[y] == r:
            out[x][y] = '1'
            entry = True
        elif grid[x][y]==1 and (r_count[x] !=c and c_count[y] != r):
            print ('NO')
            exit()
        else:
            out[x][y] = '0'
if not entry and sum(r_count)+sum(c_count)>0:
    print ('NO')
    exit()
print ('YES')
print ('\n'.join([' '.join(x) for x in out]))