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 collections import Counter,defaultdict,deque

from math import factorial as fact

def primes(n):
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
    if n > 1:
        factors.append(n)
    return factors

def issubseq(s,t):
    if len(t)>len(s):
        return False
    lt = len(t)
    p = 0
    for el in s:
        if el == t[p]:
            p+=1
        if p==lt:
            return True
    return False



#a = [int(x) for x in input().split()]

m,n = [int(x) for x in input().split()]
matrix = [[int(x) for x in input().split()] for k in range(m)]
rows = []
cols = []
res = [[0]*n for x in range(m)]
for i in range(m):
    if all(matrix[i]):
        rows.append(1)
    else:
        rows.append(0)
for i in range(n):
    ok = True
    for j in range(m):
        if matrix[j][i] == 0:
            ok = False
            break
    if ok:
        cols.append(1)
    else:
        cols.append(0)

for i in range(m):
    for j in range(n):
        res[i][j] = cols[j] and rows[i]

#print(res)
for i in range(m):
    for j in range(n):
        if matrix[i][j]:
            if any(res[i]) or any(res[x][j] for x in range(m)):
                pass
            else:
                print('NO')
                exit()
print('YES')
for el in res:
    print(*el)