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:
m, n = [int(s) for s in input().split(' ')]
B = [[] for _ in range(m)]
A = [[1 for _ in range(n)] for _ in range(m)]
ARow = [n for _ in range(m)]
ACol = [m for _ in range(n)]
for i in range(m):
    B[i] = [int(s) for s in input().split(' ')]
    for j in range(n):
        if B[i][j] == 0:
            for x in range(m):
                A[x][j] = 0
                if ACol[j] > 0:
                    ARow[x] = max(ARow[x]-1, 0)
            for y in range(n):
                A[i][y] = 0
                if ARow[i] > 0:
                    ACol[y] = max(ACol[y]-1, 0)
            ARow[i] = 0
            ACol[j] = 0
contradiction = False
for i in range(m):
    if contradiction:
        break
    for j in range(n):
        if B[i][j] == 1:
            contradiction = ARow[i] == 0 and ACol[j] == 0
        if contradiction:
            break
if contradiction:
    print('NO')
else:
    print('YES')
    for i in range(m):
        print(' '.join(map(str, A[i])))