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:
n, m = [int(i) for i in input().split()]
sum1 = [0] * n
sum2 = [0] * m
res = [[0] * m for i in range(n)]
a = 0
k = [0] * 4
flag = False
for i in range(n):
    a = [int(i2) for i2 in input().split()]
    sum1[i] = sum(a)
    for i2 in range(m):
        sum2[i2] += a[i2]
for i in range(n):
    if sum1[i] == m:
        k[2] += 1
for i in range(m):
    if sum2[i] == n:
        k[3] += 1
for i in range(n):
    if (sum1[i] > k[3]) and not(sum1[i] == 0 or sum1[i] == m) or ((k[2] * k[3] == 0) and k[2] != k[3]):
        flag = True
for i in range(m):
    if (sum2[i] > k[2]) and not(sum2[i] == 0 or sum2[i] == n) or ((k[2] * k[3] == 0) and k[2] != k[3]):
        flag = True
if flag:
    print('NO')
else:
    print('YES')
    for i in range(n):
        for i2 in range(m):
            res[i][i2] = 1
    for i in range(m):
        if sum2[i] != n:
            for i2 in range(n):
                res[i2][i] = 0
    for i in range(n):
        if sum1[i] != m:
            for i2 in range(m):
                res[i][i2] = 0
    for i in range(n):
        for i2 in range(m):
            print(res[i][i2],end=' ')
        print()