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())
y=[0]
x=[0]
flag=1
c=[1]*n
for g in xrange(m):
    xx=map(int,raw_input().split())
    y.append([0]+xx)
    x.append([0]+c)
for p in xrange(1,m+1):
    for q in xrange(1,n+1):
        if y[p][q]==0:
            x[p]=[0]*(n+1)
            for r in xrange(1,m+1):
                x[r][q]=0
for p in xrange(1,m+1):
    if flag==1:
        for q in xrange(1,n+1):
            ans=0
            if y[p][q]==1:
                for dd in xrange(1,n+1):
                    ans=ans or x[p][dd]
                for r in xrange(1,m+1):
                    ans=ans or x[r][q]
                if ans==0:
                    print "NO"
                    flag=0
                    break
if flag==1:
    print "YES"
    for p in xrange(1,m+1):
        print(" ".join(str(ff) for ff in x[p][1:]))