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=map(int,input().split())

rows0=set()

cols0=set()

row_col1=[]


for i in range(n):
    p=input().split()
    for j in range(m):
        if p[j]=='0':
            rows0.add(i)
            cols0.add(j)
        else:
            row_col1.append([i,j])

if (len(list(rows0))==n and row_col1!=[]) or (len(list(cols0))==m and row_col1!=[]) :
    print('NO')
    exit() 


for k in row_col1 :
    if k[0] in rows0 and k[1] in cols0 :
        print('NO')
        exit()

print('YES')
all0=['0' for _ in range(m)]
all1=['1' for _ in range(m)]
matrixA=[]
for i in range(n):
    if i in rows0:
        matrixA.append(all0)
    else:
        matrixA.append(all1)

for j in cols0:
    for i in range(n):
        matrixA[i][j]='0'

for i in matrixA:
    print(*i)