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:
# Uses python3
import sys
import math

def fn(m,n,k):
    a = [[1]*n]*m
    x=[]
    y=[]
    for i in range(m):
        for j in range(n):
            if k[i][j]==0:
                a[i]=[0]*n
                x.append(j)
                continue
            else:
                y.append([i,j])
        continue
    
    y1 = [0]*len(y)
    for j in x:
        for i in range(m):
            a[i][j]=0
            

    for i in y:
        if 1 in a[i[0]]:
            y1[y.index(i)]=1
        else:
            for j in range(m):
                if a[j][i[1]]==1:
                    y1[y.index(i)]=1
                    break
        
    if 0 in y1:
        print("NO")
    else:
        print("YES")    
        print('\n'.join(['\t'.join([str(cell) for cell in row]) for row in a]))           
                

if __name__ == '__main__':    
    input = sys.stdin.read()
    data = list(map(int, input.split()))
    m=data[0]
    n = data[1]
    k = [[0]*n]*m
    l=0
    for i in range(m):
        k[i] = data[l+2:l+2+n]
        l=l+n
    fn(m,n,k)