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 matrix_update(matrix,i,j):
    for row in range(len(matrix)):
        if row == i:
            matrix[i] = [0]*len(matrix[0])
            continue
        matrix[row][j] = 0




m, n = map(int,input().split())
matrix = []
ans = []
for _ in range(m):
    matrix.append(list(map(int,input().split(' '))))
    ans.append([1]*n)

for i in range(len(matrix)):
    for j in range(len(matrix[0])):
        if matrix[i][j]:
            continue
        else:
            matrix_update(ans,i,j)

flag = True
b = []
for i in range(len(ans)):
    row = []
    row_truth = any(ans[i])
    for j in range(len(ans[0])):
        row.append(any([row[j] for row in ans]) or row_truth)
    b.append(row)

if b==matrix:
    print('YES')
    for row in ans:
        for item in row:
            print(item,end=" ")
        print()
else:
    print('NO')
        


# print(matrix)
# matrix_update(ans,0,0)
# print(ans)