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 transform(a, n, m):
    row = [False for i in range(n)]
    col = [False for i in range(m)]
    mat = [[True for i in range(m)] for j in range(n)]
    for i in range(n):
        for j in range(m):
            if a[i][j] == False:
                row[i] = True
                col[j] = True
    for i in range(n):
        for j in range(m):
            if row[i] == True or col[j] == True:       
                mat[i][j] = False
    return mat

def isvlid(inp, mat, m, n):
    row = [False for i in range(n)]
    col = [False for i in range(m)]
    newinp = [[False for i in range(m)] for j in range(n)]
    for i in range(n):
        for j in range(m):
            if mat[i][j] == True:
                row[i] = True
                col[j] = True
    for i in range(n):
        for j in range(m):
            if row[i] == True or col[j] == True:       
                newinp[i][j] = True
    for i in range(n):
        for j in range(m):
            if inp[i][j] != newinp[i][j]:
                return False
    return True
line = input()
lst = line.split()
n = int(lst[0])
m = int(lst[1])
'accpet input'
inp = [[False for i in range(m)] for j in range(n)]
for i in range(n):
    line = input()
    lst = line.split()
    for j in range(m):
        inp[i][j] = True if int(lst[j]) == 1 else False

mat = transform(inp, n, m)
if isvlid(inp, mat, m, n): 
    print("YES")
    for i in range(n):
        for j in range(m):
            if mat[i][j]:
                print("1", end=" ")
            else:
                print("0", end=" ")
        print()
else:
    print("NO")