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:
m,n=map(int,input().split())
l=[]
ind=[]
out=[[1]*n for i in range(m)]
def check(b):
    for i in range(m):
        if out[i][b]==1:
            return True
    return False

for i in range(m):
    k=list(map(int,input().split()))
    for j in range(len(k)):
        if k[j]==0:
            for h in range(m):
                out[h][j]=0
            for t in range(n):
                out[i][t]=0

        if k[j]==1:
            ind.append((i,j))

flag=True    
for i in ind:
    if 1 in out[i[0]] or check(i[1]):
        pass
    else:
        print('NO')
        flag=False
        break
if flag:
    print('YES')
    for i in range(m):
        for j in range(n):
            print(out[i][j],end=' ')
        print()