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:
n,m=map(int,raw_input().split())
l=[]
l2=[]
l3=[]
for i in range(n):
   l.append(map(int,raw_input().split()))
   aux=[0]*m
   aux2=[0]*m
   l2.append(aux)
   l3.append(aux2)
def f(x,y):
    for i in range(n):
        if l[i][y]==0:
            return 0
    for i in range(m):
        if l[x][i]==0:
            return 0
    return 1
def f3(x,y):
    for i in range(n):
        l2[i][y]=1
    for i in range(m):
        l2[x][i]=1
   
ans=0
for i in range(n):
    for j in range(m):
        if l[i][j]==1 :
            if(f(i,j)==1):
                l2[i][j]=1
                l3[i][j]=1
                f3(i,j)

def pri():
    for i in l3:
        cad=''
        for j in i:
            cad+=str(j)+' '
        print cad
def f2():
    for i in range(n):
        for j in range(m):
            if(l[i][j]!=l2[i][j]):
                return 0
    return 1

if(f2()==1):
    print 'YES'
    pri()
else: print 'NO'