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 PYTHON):

Solution:
m, n = map(int, raw_input().split())
mem = []
for i in xrange(m):
    temp = list(map(int, raw_input().split()))
    mem.append(temp)
r = []
c = []

for i in xrange(m):
    for j in xrange(n):
        temp = mem[i][j]
        if temp == 0:
            if i not in r:
                r.append(i)
            if j not in c:
                c.append(j)
b = [1] * n
a = []
for i in xrange(m):
    a.append(b[:])

f = 1
for i in xrange(m):
    if f == 0:
        break
    for j in xrange(n):
        temp = mem[i][j]
        if temp == 0:
            for k in xrange(n):
                a[i][k] = 0
            for l in xrange(m):
                a[l][j] = 0
        elif (i in r) and (j in c):
            f = 0

if len(r) == m or len(c) == n:
    for i in xrange(m):
        if 1 in mem[i]:
            f = 0
p = ''
if f:
    print 'YES'
    for i in xrange(m):
        for j in xrange(n):
            p += str(a[i][j])
            if len(p) == n:
                print ' '.join(p)
                p = ''
else:
    print 'NO'