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:
def solve():
    m,n = map(int,raw_input().split())
    B = []
    for i in range(m):
        B.append(map(int, raw_input().split()))
    A = [[1 for i in range(n)] for i in range(m)]
    for i in range(m):
        for j in range(n):
            if B[i][j] == 0:
                for c in range(n):
                    A[i][c] = 0
                for r in range(m):
                    A[r][j] = 0
    for i in range(m):
        for j in range(n):
            if B[i][j] == 1 and \
                    (any(x == 1 for x in [A[i][c] for c in range(n)]) or \
                     any(x == 1 for x in [A[r][j] for r in range(m)])):
                continue
            elif B[i][j] == 0 and \
                    (all(x == 0 for x in [A[i][c] for c in range(n)]) and \
                     all(x == 0 for x in [A[r][j] for r in range(m)])):
                continue
            else:
                print "NO"
                return
    print "YES"
    for i in range(m):
        print " ".join(map(str,A[i]))

solve()