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 = list(map(int, input().split()))
b = []
a = []
new_b = []
for _ in range(m):
    b.append(list(map(int, input().split())))
    a.append([1]*n)
    new_b.append([0]*n)

for i, row in enumerate(b):
    for j, col in enumerate(row):
        if col == 0:
            for c in range(n):
                a[i][c] = 0
            for r in range(m):
                a[r][j] = 0
# print(a)
for i, row in enumerate(b):
    for j, col in enumerate(row):
        flag = 0
        for c in range(n):
            if a[i][c] ==1:
                flag = 1
                break
        if flag==0:
            for r in range(m):
                if a[r][j] ==1:
                    flag = 1
                    break
        new_b[i][j] = flag
if new_b == b:
    print("YES")
    for row in a:
        for item in row:
            print(item,  end=" ")
        print()
else:
    print("NO")