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:
inp = raw_input()
n, m = inp.split(" ")
n = int(n)
m = int(m)
mat = {}
for i in xrange(n):
    row = raw_input()
    elems = row.split(" ")
    for j in xrange(m):
        mat[(i, j)] = int(elems[j])

res = {}
for i in xrange(n):
    for j in xrange(m):
        all1 = True
        for k in xrange(m):
            if mat[(i, k)] != 1:
                all1 = False
                break
        for k2 in xrange(n):
            if mat[(k2, j)] != 1:
                all1 = False
                break
        if all1:
            res[(i,j)] = 1
        else:
            res[(i,j)] = 0


isValid = True
for i in xrange(n):
    if not isValid:
        break
    for j in xrange(m):
        if not isValid:
            break
        all0 = True
        for k in xrange(m):
            if res[(i, k)] != 0:
                all0 = False
                break
        for k2 in xrange(n):
            if res[(k2, j)] != 0:
                all0 = False
                break

        if all0 and mat[(i,j)] == 1:
            isValid = False
        if not all0 and mat[(i, j)] == 0:
            isValid = False

if not isValid:
    print "NO"
else:
    print "YES"
    for i in xrange(n):
        ans = []
        for j in xrange(m):
            ans.append(str(res[(i, j)]))
        print " ".join(ans)