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:
import sys

# sys.stdin = open("7.txt", "r")

m, n = map(int, input().split())

oneUps = [0 for _ in range(n)]

oneRights = [0 for _ in range(m)]

zeroUps = set()

zeroRights = set()

B = []

cnt = 0

for i in range(m):
    row = list(map(int, input().split()))
    B.append(row)
    for j in range(n):
        if row[j] == 1:
            oneUps[j] += 1
            oneRights[i] += 1
            cnt += 1
        else:
            zeroUps.add(j)
            zeroRights.add(i)


Ones = set()

i = 0
j = 0
for i in range(m):
    for j in range(n):
        if oneUps[j] == m and oneRights[i] == n:
            Ones.add((i, j))

grid = [[None for _ in range(n)] for _ in range(m)]

i = 0
j = 0

result = True

for i in range(m):
    for j in range(n):
        
        if (i, j) in Ones:
            if B[i][j] == 0:
                result = False
            grid[i][j] = 1

        if i in zeroRights or j in zeroUps:
            grid[i][j] = 0

        if (i, j) in Ones and (i in zeroRights or j in zeroUps):
            result = False


        if B[i][j] == 1:
            if not oneUps[j] == m and  not oneRights[i] == n:
                result = False


i = 0
j = 0

if len(Ones) == 0 and cnt >= 1:
    print("NO")
else:
    if result:
        print("YES")
        for i in range(m):
            print(" ".join(map(str, grid[i])))
    else:
        print("NO")