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:
#B problem
X = raw_input()
X = X.split()
#print X
m = int(X[0])
n = int(X[1])
A = [[0]*n for i in range(m)]
#print A
B = [[0]*n for i in range(m)]
for i in range(m):
    temp = raw_input()
    for j in range(n):
        B[i][j] = int(temp[2*j])
temp1 = []
temp2 = []
for i in range(m):
    if (sum(B[i]) == n):
        temp1.append(i)
for j in range(n):
    if (sum([row[j] for row in B]) == m):
        temp2.append(j)
if((len(temp1) != 0) & (len(temp2) != 0)):
    for i in range(len(temp1)):
        for j in range(len(temp2)):
            A[temp1[i]][temp2[j]] = 1
count1 = 0
count2 = 0
for i in range(m):
    for j in range(n):
        if (B[i][j] == 1):
            count1 += 1
for i in range(m):
    for j in range(n):
        if (B[i][j] == 0):
            count2 += 1
#print temp1, temp2, len(temp1), len(temp2), m, n
if(count2 == (m*n)):
    print "YES"
    C = [[0]*n for i in range(m)]
    for i in range(m):
        for j in range(n):
            print C[i][j],
            print "\n",
elif((count1 == (len(temp1)*n + len(temp2)*m - len(temp1)*len(temp2))) & ((len(temp1) != 0) & (len(temp2) != 0))):
    print "YES"
    for i in range(m):
        for j in range(n):
            print A[i][j],
            print "\n",
else:
    print "NO"