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 PYTHON3):

Solution:
m,n=map(int,input().split())
s=[]
t=[]
e=[]
y=0
z=0
for i in range(m):
    d=list(map(int,input().split()))
    s.append(d)
    t.append([1]*len(d))
for i in range(len(s)):
    for j in range(len(s[i])):
        if s[i][j]==0:
            t[i]=[0]*n
            for k in t:
                k[j]=0
for i in range(m):
    q=[]
    for j in range(n):
        p=t[i][j]
        for z in range(n):
            p=p or t[i][z]
        if p==0:
            for v in range(m):
                p=p or t[v][j]
        q.append(p)
    e.append(q)
if  e!=s:
    print ('NO')
else:
    print('YES')
    for i in t:
        for j in i:
            print(j,end=' ')
        else:
            print( )