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 find_matrix(s,matx):

    nbL=s[0] # nombre de lignes
    nbC=s[1] #nombre de colonnes
    
    trouvee = [[1 for i in range(nbC)] for j in range(nbL)]
    recree = [[0 for i in range(nbC)] for j in range(nbL)]
    alreadychecked = [[False for i in range(nbC)] for j in range(nbL)]
    
    for l in range(nbL): #RE-CREATION MATRICIELLE
        for c in range(nbC):
            
            if matx[l][c]==0:
                if not alreadychecked[l][c] :
                    for l2 in range(nbL):
                        trouvee[l2][c]=0
                        
                    for c2 in range(nbC):
                        trouvee[l][c2]=0
                        
                        alreadychecked[l][c]=True

    for l in range(nbL):
        for c in range(nbC):
            if trouvee[l][c]==1:
                
                for k in range(nbL):
                    recree[k][c]=1
                for k in range(nbC):
                    recree[l][k]=1
                    
    if recree==matx:
        print("YES")
        for i in range(nbL):
            for j in range(nbC-1):
                print(trouvee[i][j], end=" ")
            print(trouvee[i][nbC-1])

    else:
        print("NO")
        
    
    
s = list(map(int, input().split()))
matx = [list(map(int, input().split())) for line in range(s[0])]

find_matrix(s, matx)