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:
inp = raw_input().split()

m = int(inp[0])
n = int(inp[1])
unit_col = set()
unit_str = set()
zero_col = set()
B = [['1'] * n for i in range(m)]
zero_ind = 0
exit = 0

for i in range(m):
    inp = raw_input().split()
    if '1' not in inp:
        zero_ind = 1
    elif '0' not in inp:
        unit_str.add(i)
    else:
        for j in range(n):
            #print(i),
            #print(j),
            #print(inp[j])
            B[i][j] = inp[j]
            if inp[j] == '1':
                unit_col.add(j)
            else:
                zero_col.add(j)

#print(unit_str)
#print(unit_col)

#print "B ",
#print B

if zero_ind == 1:
    if ((len(unit_str) > 0) or (len(unit_col) > 0)):
        print("NO")
        exit = 1
    else:
        print("YES")
else:
    if (len(unit_str) == 0):
        print("NO")
        exit = 1
    else:
        #print unit_col
        for i in unit_col:
            #print(B[:][i])
            #print list(B[:][i:i+1])
            if i in zero_col:
                #print i,
                #print(B[:][i:i+1]), 
                print("NO")
                exit = 1
                break
        if (exit == 0):
            print("YES")
            
if (exit == 0):
    for i in range(m):
        for j in range(n):
            if ((j in unit_col) or (len(unit_col) == 0)) and (i in unit_str):
                print('1'),
            else:
                print('0'),
        print('')