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:
from sys import stdin


def main():
    m, n = map(int, stdin.readline().split())
    b = []
    for _ in range(m):
        b.append(list(map(int, stdin.readline().split())))
    a = [[0 for _ in range(n)] for _ in range(m)]
    for i in range(m):
        for j in range(n):
            if all(b[i][k] == 1 for k in range(n)) and all(b[k][j] == 1 for k in range(m)):
                a[i][j] = 1
    ok = True
    for i in range(m):
        for j in range(n):
            curr = 0
            for k in range(n):
                curr = curr or a[i][k]
            for k in range(m):
                curr = curr or a[k][j]
            if curr != b[i][j]:
                ok = False
                break
    if not ok:
        print("NO")
    else:
        print("YES")
        for r in a:
            print(*r)


if __name__ == "__main__":
    main()