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 fn(a, i, j, m, n):
    x, y = i, j

    while (x > 0):
        x -= 1
        if(a[x][y] != 1):
            return 0
    x, y = i, j

    while(x < m - 1):
        x += 1
        if(a[x][y] != 1):
            return 0

    x, y = i, j

    while(y > 0):
        y -= 1
        if(a[x][y] != 1):
            return 0
    x, y = i, j

    while(y < n - 1):
        y += 1
        if(a[x][y] != 1):
            return 0

    return 1


m, n = map(int, input().split())
b = []

for i in range(m):
    p = list(map(int, input().split()))
    b.append(p[:])

d = []
for i in range(m):

    for j in range(n):

        if(b[i][j] == 1):
            k = fn(b, i, j, m, n)
            if(k == 1):
                d.append([i, j])

# print(d)
a = []
c = []
p = [0] * n
for i in range(m):
    a.append(p[:])
    c.append(p[:])
for u, v in d:
    a[u][v] = 1

    for i in range(m):
        c[i][v] = 1
    for j in range(n):
        c[u][j] = 1
# print(b)
# print(c)
f = 0
for i in range(m):
    for j in range(n):
        if(c[i][j] != b[i][j]):
            f = 1
            break
if(f == 1):
    print("NO")
else:
    print("YES")
    for i in range(m):
        print(*a[i])