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:
from array import *

m, n = list(map(int, input().split()))

A = []

for i in range(m):
    k = list(map(int, input().split()))
    A.append(k)

ans = [[-1 for x in range(n)] for y in range(m)]

flag = 0

for i in range(m):
    if flag:
        break
    for j in range(n):
        if A[i][j] == 0:
            # print()
            # print(i, j)
            for a in range(n):
                # print(i, a)
                if ans[i][a] == -1 or ans[i][a] == 0:
                    # print(i, a)
                    # print(ans)
                    ans[i][a] = 0
                    # print(ans)
                    # print()
                else:
                    print("NO")
                    flag = 1
                    break
            if flag:
                break
            for b in range(m):
                if ans[b][j] == -1 or ans[b][j] == 0:
                    # print(b, j)
                    # print(ans)
                    ans[b][j] = 0
                    # print(ans)
                    # print()
                    # print(ans)
                else:
                    print("NO")
                    flag = 1
                    break

for i in range(m):
    if flag:
        break
    for j in range(n):
        if ans[i][j] == -1:
            if A[i][j] == 1:
                ans[i][j] = 1
            else:
                print("NO")
                flag = 1
                break

""""""

for i in range(m):
    check = 0
    if flag:
        break
    for j in range(n):
        check = 0
        if A[i][j] == 1:
            #print()
            #print(i,j)
            for a in range(n):
                if ans[i][a] == 1:
                    check = 1
                    #print("check = 1 : " + str(i) + " , " + str(a))
            for b in range(m):
                if ans[b][j] == 1:
                    check = 1
                    #print("check = 1" + str(b) +" , "+ str(j))
            if check:
                continue
            else:
                flag = 1
                print("NO")
                break



if flag == 0:
    print("YES")
    for i in ans:
        for j in i:
            print(j, end=" ")
        print()

"""
2 3
0 1 0
1 1 1
"""