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:
"""
    Template written to be used by Python Programmers.
    Use at your own risk!!!!
    Owned by adi0311(rating - 1989 at CodeChef and 1449 at Codeforces).
"""
import sys
import bisect
import heapq
from math import *
from collections import defaultdict as dd  # defaultdict(<datatype>) Free of KeyError.
from collections import deque  # deque(list) append(), appendleft(), pop(), popleft() - O(1)
from collections import Counter as c  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb
from bisect import bisect_left as bl, bisect_right as br, bisect
# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(var)
def l(): return list(map(int, data().split()))
def sl(): return list(map(str, data().split()))
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [[val for i in range(n)] for j in range(m)]


m, n = sp()
b = list()
for i in range(m):
    b.append(l())
a = l2d(n, m, 1)
r, c = set(), set()
for i in range(m):
    for j in range(n):
        if b[i][j] == 0:
            r.add(i)
            c.add(j)
for i in r:
    for j in range(n):
        a[i][j] = 0
for i in c:
    for j in range(m):
        a[j][i] = 0
rowor = list()
for i in range(m):
    temp = 0
    for j in range(n):
        temp = temp | a[i][j]
        if temp == 1:
            break
    rowor.append(temp)
color = list()
for i in range(n):
    temp = 0
    for j in range(m):
        temp = temp | a[j][i]
        if temp == 1:
            break
    color.append(temp)
result = True
for i in range(m):
    for j in range(n):
        temp = rowor[i] | color[j]
        if temp != b[i][j]:
            result = False
    if not result:
        break
if result:
    print("YES")
    for i in a:
        print(*i)
    exit()
print("NO")