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:
l=lambda:map(int,raw_input().split())

n,m=l()
M=[l() for i in range(n)]
N=[[0]*m for i in range(n)]
S=0
F=0
T=1
for i in range(n):
    for j in range(m):

        N[i][j] = str(int(all(M[i]) and all([M[k][j] for k in range(n)])))
        S +=all([M[k][j] for k in range(n)])
        F +=all(M[i])
        if M[i][j]:
            T = T and (all(M[i]) or all([M[k][j] for k in range(n)]))
#print S,F,bool(S)^bool(F)
if T and not (bool(S)^bool(F)):
    print 'YES'
    print '\n'.join(map(lambda s: ' '.join(s), N))
else:
    print 'NO'