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:
def or_matrix(B):
    m = len(B)
    n = len(B[0])
    A = [[1 for i in range(n)] for j in range(m)]
    C = [[0 for i in range(n)] for j in range(m)]
    s_r = set()
    s_c = set()
    for i in range(len(B)):
        for j in range(len(B[0])):
            if B[i][j] == 0:
                s_r.add(i)
                s_c.add(j)

    for i in range(len(A)):
        for j in range(len(A[0])):
            if i in s_r:
                A[i][j] = 0
            if j in s_c:
                A[i][j] = 0
    
    n_s_r = set()
    n_s_c = set()
    for i in range(len(A)):
        for j in range(len(A[0])):
            if A[i][j] == 1:
                n_s_r.add(i)
                n_s_c.add(j)
                
                
    for i in range(len(C)):
        for j in range(len(C[0])):
            if i in n_s_r:
                C[i][j] = 1
            if j in n_s_c:
                C[i][j] = 1
    if B == C:
        print('YES')
        for i in range(len(A)):
            for j in range(len(A[0])):
                print(A[i][j],end=' ')
            print()
    else:
        print('NO')
    

m,n = [int(x) for x in input().split()]

B = [[int(j) for j in input().split()] for i in range(m)]
or_matrix(B)