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:
#sys.setrecursionlimit(300000)
import sys
def main():
    pass
def binary(n):
    #decimal to binary
    return (bin(n).replace("0b", ""))
def decimal(s):
    #binary to decimal
    return (int(s, 2))
def pow2(n):
    #power of a number base 2
    p = 0
    while n > 1:
        n //= 2
        p += 1
    return (p)
def isPrime(n):
    # if  number is prime in √n time
    if (n == 1):
        return (False)
    else:
        root = int(n ** 0.5)
        root += 1
        for i in range(2, root):
            if (n % i == 0):
                return (False)
        return (True)
def lts(l):
    #list to string ,no spaces
    s=''.join(map(str,l))
    return s
def stl(s):
    #for each character in string to list with no spaces -->
    l=list(s)
    #for space in string  -->
    #l=list(s.split(" "))
    return l
#Returns list of numbers with a particular sum
def sq(a,target,arr=[]):
    s=sum(arr)
    if(s==target):
        return arr
    if(s>=target):
        return
    for i in range(len(a)):
        n=a[i]
        remaining=a[i+1:]
        ans=sq(remaining,target,arr+[n])
        if(ans):
            return ans

#1000000007
mod=int(1e9)+7
def sinp(): return sys.stdin.readline().strip()
def iinp(): return int(input())
def ninp(): return map(int, sys.stdin.readline().strip().split())
def llinp(): return list(map(int, sys.stdin.readline().strip().split()))
def p(xyz): print(xyz)
def p2(a,b):print(a,b)
import math as m
#for _ in range(iinp()):
m,n=ninp()
ans=[[1 for i in range(n)] for j in range(m)]
mat=[]
for i in range(0,m):
    a=llinp()
    mat.append(a)
for i in range(0,m):
    for j in range(0,n):
        if(mat[i][j]==0):
            for x in range(0,n):
                ans[i][x]=0
            for x in range(0,m):
                ans[x][j]=0
for i in range(0,m):
    for j in range(0,n):
        f1 = 0
        f2 = 0
        if(mat[i][j]==1):
            if(1 in ans[i]):
                f1=1
            for x in range(m):
                if(ans[x][j]==1):
                    f2=1
            if(f1==1 or f2==1):
                pass
            else:
                print("NO")
                exit()
print("YES")
for i in ans:
    print(*i)