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())
row=[1]*m
col=[1]*n
cache=[]
for i in range(m):
    s=map(int,raw_input().split())
    cache.append(s)
    for j in range(n):
        if s[j]==0:
            row[i]=0
            col[j]=0
ans=[[0 for i in range(n)] for j in range(m)]
for i in range(m):
    for j in range(n):
        if row[i]==1 and col[j]==1:
            ans[i][j]=1
check=True
for i in range(m):
    for j in range(n):
        if cache[i][j]==1 and row[i]==0 and col[j]==0:
            check=False
            break
nrow=[0]*m
ncol=[0]*n
for i in range(m):
    for j in range(n):
        if ans[i][j]==1:
            nrow[i]=1
            ncol[j]=1
for i in range(m):
    if nrow[i]!=row[i]:
        check=False
        break
for j in range(n):
    if ncol[j]!=col[j]:
        check=False
        break
if check:
    print 'YES'
    for i in range(m):
        for j in range(n):
            print ans[i][j],
        print '\n'
else:
    print 'NO'