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:
x,y=map(int,input().split())
lol=[]
for __ in range(x):
    lol.append(list(map(int,input().split())))
yo=[]
for i in lol:
    po=i[:]
    yo.append(po)
for i in range(x):
    k=1
    for j in range(y):
        k&=lol[i][j]
    for j in range(y):
        yo[i][j]&=k
    #print(yo,k,lol)    
for i in range(y):
    k=1
    for j in range(x):
        k&=lol[j][i]
    for j in range(x):
        yo[j][i]&=k

#print(yo,lol)      

yo0=[]
for i in yo:
    po=i[:]
    yo0.append(po)
for i in range(x):
    k=0
    for j in range(y):
        k|=yo[i][j]
    for j in range(y):
        yo0[i][j]|=k
for i in range(y):
    k=0
    for j in range(x):
        k|=yo[j][i]
    for j in range(x):
        yo0[j][i]|=k
if yo0==lol:
    print("YES")
    for i in yo:
        print(*i)
else:
    print("NO")
        

#0 in a row then whole row 0, implement it afterwards this implementation