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:
from collections import deque, Counter, OrderedDict
from heapq import nsmallest, nlargest
from math import ceil,floor,log,log2,sqrt,gcd,factorial,pow,pi
from bisect import bisect_left,bisect_right

def binNumber(n,size=4):
    return bin(n)[2:].zfill(size)

def iar():
    return list(map(int,input().split()))

def ini():
    return int(input())

def isp():
    return map(int,input().split())

def sti():
    return str(input())

def par(a):
    print(' '.join(list(map(str,a))))

def tdl(outerListSize,innerListSize,defaultValue = 0):
    return [[defaultValue]*innerListSize for i in range(outerListSize)]

def sts(s):
    s = list(s)
    s.sort()
    return ''.join(s)

class pair:
    def __init__(self,f,s):
        self.fi = f
        self.se = s
    def __lt__(self,other):
        return (self.fi,self.se) < (other.fi,other.se)

#  =========     /\       /|    |====/|
#      |        /  \       |    |   / |
#      |       /____\      |    |  /  |
#      |      /      \     |    | /   |
#  ========= /        \  =====  |/====|  
#  code

def check(a):
    s = a[0][0]+a[1][0]+a[2][0]
    for i in range(3):
        if sum(a[i]) != s:
            return False
    for i in range(3):
        s1 = 0
        for j in range(3):
            s1 += a[i][j]
        if s1 != s:
            return False
    
    s1 = a[1][1] + a[2][2] + a[0][0]
    s2 = a[2][0] + a[1][1] + a[0][2]
    
    if s1 != s or s2 != s:
        return False
    
    return True

if __name__ == "__main__":
    n,m = isp()
    a = tdl(n,m,1)
    b = tdl(n,m)
    b1 = tdl(n,m)

    r = [0]*n
    c = [0]*m

    for i in range(n):
        b[i] = iar()

    for i in range(n):
        for j in range(m):
            if b[i][j] == 0:
                a[i] = [0]*m
                for k in range(n):
                    a[k][j] = 0
    

    for i in range(n):
        for j in range(m):
            r[i] = r[i] or a[i][j]
    
    for i in range(m):
        for j in range(n):
            c[i] = c[i] or a[j][i]

    for i in range(n):
        for j in range(m):
            b1[i][j] = r[i] or c[j]
    
    if b1 == b:
        print('YES')
        for i in range(n):
            print(' '.join(list(map(str,a[i]))))
    else:
        print('NO')