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:
from sys import stdin
n, m = map(int, stdin.readline().rstrip().split(" "))
l = []
for _ in range(n):
    al = list(map(int, stdin.readline().rstrip().split(" ")))
    l.append(al)

re =" YES"
sol = [[0]*m for i in range(n)]

for i in range(n):
    for j in range(m):
        if l[i][j]==1:
            c1 = True
            c2 = True
            for k in range(n):
                if l[k][j]==0:
                    c1 = False
            for o in range(m):
                if l[i][o] == 0:
                    c2 = False

            if c1 and c2:
                sol[i][j] = 1
            elif not c1 and not c2:
                re = "NO"
                break

if re =="NO":
    print("NO")
else:
    test = [[0]*m for i in range(n)]
    for i in range(n):
        for j in range(m):
            if sol[i][j]==1:
                for k in range(n):
                    test[k][j] =1

                for o in range(m):
                    test[i][o] = 1
    if test == l:
        print("YES")
        for i in range(n):
            print(*sol[i])
    else:
        print("NO")