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 PYTHON):

Solution:
 # -*- coding: utf-8 -*- 
import sys
mn=raw_input().split()
m=int(mn[0])
n=int(mn[1])

B= []
A = [[-1 for i in range(n)] for j in range(m)]
for i in range(m):
    row = [int(r) for r in raw_input().split()]
    B.append(row)

    
for i in range(m):
    for j in range(n):
        if B[i][j]==0:
            for col in range(n):
                if A[i][col] != 1:
                    A[i][col]=0
                else:
                    print "NO"
                    sys.exit()
            for row in range(m):
                if A[row][j]!=1:
                    A[row][j]=0                    
                else:
                    print "NO"
                    sys.exit()

for i in range(m):
    for j in range(n):
        if A[i][j]==-1:
            row = A[i][:]
            row[j]=1
            A[i]=row
C=[]
for i in range(m):
    row = []
    for j in range(n):
        col=[]
        for c in range(m):
            col.append(A[c][j])
        redrow = reduce(lambda x, y: x or y, A[i][:])
        redcol = reduce(lambda x, y: x or y, col)
        row.append(redrow or redcol)
    C.append(row)

for i in range(m):
    if C[i]!=B[i]:
        print "NO"
        sys.exit()

print "YES"
for i in range(m):
    print reduce(lambda x, y: str(x) + ' ' + str(y), A[i])