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:
#br = open('b.in')

def solve(n, m):
  a = [f() for i in range(n)]
  b = [[0 for i in range(m)] for j in range(n)]
  zn = [i for j in range(m) for i in range(n) if a[i][j] == 0 ]
  zm = [j for j in range(m) for i in range(n) if a[i][j] == 0 ]

  for i in range(n):
    for j in range(m):
      if a[i][j] == 1 and i not in zn and j not in zm:
        b[i][j] = 1

  for i in range(n):
    for j in range(m):
      if (a[i][j] == 0) and (sum([b[x][j] for x in range(n)]) + sum([b[i][y] for y in range(m)]) > 0):
        return None 
      if a[i][j] == 1 and sum([b[x][j] for x in range(n)]) + sum([b[i][y] for y in range(m)]) <= 0:
        return None 

  return b

f = lambda: map(int, raw_input().split())
n, m = f()
a = solve(n, m)
if a is None:
  print "NO"
else:
  print "YES"
  for i in range(n):
    print " ".join(map(str, a[i]))