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:
__author__ = '-'

l1 = input().strip().split(" ")
m = int(l1[0])
n = int(l1[1])

B = []
for i in range(m):
    line = input().strip().split(" ")
    row = []
    for p in line:
        row.append(int(p))
    B.append(row)

A = [[0 for x in range(n)] for y in range(m)]

def transferA(B,m,n):
    A = [[0 for x in range(n)] for y in range(m)]

    for i in range(m):
        rida1 = True
        for j in range(n):
            if B[i][j] == 0:
                rida1 = False
                break
        if rida1:
            for j in range(n):
                veerg1 = True
                for r in range(m):
                    if B[r][j] == 0:
                        veerg1 = False
                        break
                if veerg1 and rida1:
                    A[i][j] = 1
    return A

A = transferA(B,m,n)
def transfer(A,m,n):
    C = [[0 for x in range(n)] for y in range(m)]
    for i in range(len(A)):
        for j in range(len(A[i])):
            if A[i][j] == 1:
                for k in range(n):
                    C[i][k] = 1
                for l in range(m):
                    C[l][j] = 1
    return C


C = transfer(A,m,n)

if C == B:
    print("YES")
    for i in range(len(A)):
        for j in range(len(A[i])):
            print(A[i][j],end = " ")
        print()
else:
    print("NO")