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:
'''
Created on Sep 14, 2015

@author: sshadmin
'''
m,n=map(int,raw_input().strip().split())

import sys
grid=[]
for i in range(m):
    grid.append(map(int,raw_input().strip().split()))
r,c=set([]),set([])
for i in range(m):
    if all([True if grid[i][x] ==1 else 0 for x in range(n)]):
        r.add(i)
for i in range(n):
    if all([True if grid[x][i] ==1 else 0 for x in range(m)]):
        c.add(i)   
if sum([sum(grid[x]) for x in range(m)])==0:
    print "YES"
    for i in range(m):
        print " ".join(["0"]*n)
elif len(r) ==0 or len(c)==0:
    print "NO"
else:
    for i in range(m):
        for j in range(n):
            if grid[i][j]==1 and i not in r and j not in c:
                print "NO"
                sys.exit()
    rgrid=[]
    for i in range(m):
        rgrid.append([0]*n)
    r,c=list(r),list(c)
    for i in range(len(r)):
        for j in range(len(c)):
            rgrid[r[i]][c[j]]=1
    print "YES"
    for i in range(m):
        print " ".join(map(str,rgrid[i]))