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 main():
    (n, m) = map(int, input().split(' '))
    B = []
    for i in range(n):
        B.append(list(map(int, input().split(' '))))
    r = []
    c = []
    for i in range(n):
        if len(list(filter(lambda x: x == 1, B[i]))) == len(B[i]):
            r.append(1)
        else:
            r.append(0)
    for j in range(m):
        col = [B[i][j] for i in range(n)]
        if len(list(filter(lambda x: x == 1, col))) == len(col):
            c.append(1)
        else:
            c.append(0)

    A = [[-1 for i in range(m)] for j in range(n)]
    yes = True
    for i in range(n):
        for j in range(m):
            if r[i] == 1 and c[j] == 1:
                A[i][j] = 1
            else:
                A[i][j] = 0
    for i in range(n):
        for j in range(m):
            o = 0
            for k in range(n):
                o |= A[k][j]
            for k in range(m):
                o |= A[i][k]
            if o != B[i][j]:
                print("NO")
                return
    print("YES")
    for i in range(n):
        print(' '.join(list(map(str, A[i]))))
main()