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:
# cook your dish here
m,n=map(int,input().split())
l=[]
for i in range(m) :
    l.append(list(map(int,input().split())))
r=[]
c=[]
f=0
for i in range(m) :
    for j in range(n) :
        if l[i][j]==1 :
            x=1 
            for k in range(m) :
                if l[k][j]==0 :
                    x=0 
                    break
            if x==1 :
                c.append(j)
                for k in range(n) :
                    if l[i][k]==0 :
                        x=0 
                        break
                if x==1 :
                    r.append(i)
            else :
                for k in range(n) :
                    if l[i][k]==0 :
                        x=1 
                        break
                if x==0 :
                    r.append(i)
                else :
                    f=1 
                    break
if f==1 or (len(r)==0 and len(c)!=0) or (len(c)==0 and len(r)!=0):
    print("NO")
else :
    print("YES")
    for i in range(m) :
        for j in range(n) :
            if i in r and j in c :
                print(1,end=" ") 
            else :
                print(0,end=" ")
        print()