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 PYTHON):

Solution:
row,column=map(int,raw_input().split())
matrix=[]
for i in range(row):
    matrix.append(map(int,raw_input().split()))
    
test=[[1 for _ in range(column)] for _ in range(row)]

def erase(r,c):
    
    for i in range(column):
        test[r][i]=0
    for i in range(row):
        test[i][c]=0

def or_matrix(ma):
    mb=[[0 for _ in range(column)] for _ in range(row)]
    for i in range(row):
        for j in range(column):
            one=False
            for k in range(column):
                if ma[i][k]==1:
                    one=True
            for k in range(row):
                if ma[k][j]==1:
                    one=True
            if one:
                mb[i][j]=1
    
    return mb


for dummy_row in range(row):
    for dummy_column in range(column):
        if matrix[dummy_row][dummy_column]==0:
            erase(dummy_row,dummy_column)
            
m=or_matrix(test)
if m==matrix:
    print 'YES'
    for i in range(row):
        for j in test[i]:
            print j,
        print
else:
    print 'NO'