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:
import sys
def get_ints(): return list(map(int, sys.stdin.readline().strip().split()))
 
temparr = get_ints()
r = int(temparr[0])
c = int(temparr[1])
arr = []
rowfull = []
flag = 0 
temparr = [ -1 for _ in range(c) ]
ans  = [temparr[::] for _ in range(r)]
for i in range(r):
    newinput = get_ints()
    arr.append(newinput[::])
    for j in range(c):
        if newinput[j] == 0 :
            for k in range(c):
                ans[i][k] = 0
            for k in range(r):
                ans[k][j] = 0
flag = 0 
for i in range(r):
    for j in range(c):
        flag = 0 
        # if i == (r-1) and j == (c - 1):
        #     for i in range(r):
        #         strs = ""
        #         for j in range(c):
        #             strs += str(ans[i][j])
        #             strs += " "
        #         print(strs)
        if arr[i][j] == 1 :
            #print(str(i) + " " + str(j) )
            for k in range(c):
                if ans[i][k] == -1 or ans[i][k] == 1:
                    ans[i][k]  = 1
                    flag = 1 
            for k in range(r):
                if ans[k][j] == -1 or ans[k][j] == 1:
                    ans[k][j] = 1 
                    flag = 1 
            
            if flag == 0 :
                flag = 2 
                break
        if flag == 2:
            break 
    if flag == 2:
        break 
if flag == 2:
    print("NO")
else:
    print("YES")
    for i in range(r):
        strs = ""
        for j in range(c):
            strs += str(ans[i][j])
            strs += " "
        print(strs)