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 main():
    n,m  = map(int,input().split())
    a = [[1 for i in range(m)] for __ in range(n)]
    ones = []
    zer_r = [1 for i in range(n)]
    zer_c = [1 for j in range(m)] 
    for i in range(n):
        b = list(map(int,input().split()))
        for j in range(m):
            if b[j]==0:
                zer_r[i] = 0
                zer_c[j] = 0
            else:
                ones.append((i,j))
    poss = True
    for i in ones:
        if zer_r[i[0]] == 0 and zer_c[i[1]]==0:
            poss = False
    if (sum(zer_r) == 0 and sum(zer_c) >= 1) or (sum(zer_c) == 0 and sum(zer_r) >= 1):
        poss = False 
    
    if poss:
        for i in range(n):
            for j in range(m):
                if zer_r[i] and zer_c[j]:
                    a[i][j]=1
                else:
                    a[i][j]=0
        
        print('YES')
        for i in a:
            print(*i)
    else:
        print("NO")
if __name__ == "__main__":
    main()