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:
def set(ma,i,j,m,n,f):
    for k in xrange(n):
        ma[i][k]=f
    for k in xrange(m):
        ma[k][j]=f
    return ma

def check(ma,mb,m,n):
    for i in xrange(m):
        for j in xrange(n):
            if ma[i][j]!=mb[i][j]:
                return False
    return True
def run():
    m,n=map(int,raw_input().split(" "))
    ma=[ [1 for j in xrange(n)] for i in xrange(m)]
    ob=[]
    for i in xrange(m):
        nline=map(int,raw_input().split(" "))
        ob.append(nline)
        for j in xrange(n): 
            if nline[j]==0:
                ma=set(ma,i,j,m,n,0)
    # re-gen b
    mb=[ [0 for j in xrange(n)] for i in xrange(m)]
    for i in xrange(m):
        for j in xrange(n):
            if ma[i][j]==1:
                mb=set(mb,i,j,m,n,1)
    if check(mb,ob,m,n) :
        print "YES"
        for i in xrange(m):
            print " ".join(map(str,ma[i]))
    else:
        print "NO"

run()