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:
m,n=list(map(int,input().split()))
grid = [list(map(int,input().split())) for i in range(m)]
def ANDrow(grid,i):
    ans = 1
    for j in range(n):
        ans = ans and grid[i][j]
    return ans
def ANDcol(grid,j):
    ans = 1
    for i in range(m):
        ans = ans and grid[i][j]
    return ans
gridA = []
for i in range(m):
    temp= []
    for j in range(n):
        temp.append(0)
    gridA.append(temp)
# gridA= [[0]*n]*m
flag = True
countR = 0
countC = 0
for i in range(m):
    if not flag:
        break
    for j in range(n):
        androw = ANDrow(grid[::],i)
        countR+=androw
        andcol = ANDcol(grid[::],j)
        countC+=andcol
        # print(grid[i][j],androw,andcol,grid[i][j] and androw and andcol)
        gridA[i][j] = int(bool(grid[i][j]) and bool(androw) and bool(andcol))
        # print(i,j,int(bool(grid[i][j]) and bool(androw) and bool(andcol)),gridA)
        flag2 = (grid[i][j] and (not (androw or andcol)))
        if flag2:
            flag = False
            break
if flag and not((countC==0) ^ (countR==0)):
    print("YES")
    for i in gridA:
        print(*i)
else:
    print("NO")