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:
def OrMat():
    M,N = input().split( )
    m = int(M)
    n = int(N)
    A = []
    for _ in range(m):
        ip_arr = [int(i) for i in input().split(" ")]
        A.append(ip_arr)
    
    # check left top right and bootom of matrix index even one of index which of them contains 0 then B[i][j] surely have to 0
    B = []
    for _ in range(m):
        temp = []
        for _ in range(n):
            temp.append(-1)
        B.append(temp)  
    for i in range(m):        
        for j in range(n):
            p = i
            q = 0
            # row traversal
            while q < n:
                if A[p][q] == 0:
                    B[i][j] = 0
                q+=1
            
            p = 0
            q = j
            # col traversal
            while p < m:
                if A[p][q] == 0:
                    B[i][j] = 0
                p+=1
            if B[i][j] == -1:
                B[i][j] = 1
            else:
                B[i][j] = 0
    # check  whether matrix B[i][j] in i row or j col or operation result is same as A[i][j] or not i if same then Print Yes otherwise print No
    res = 0
    for i in range(m):
        for j in range(n):
            p = i
            q = 0
            while q < n:
                res = res or B[p][q]
                q+=1
            
            p = 0
            q = j
            while p < m:
                res = res or B[p][q]
                p+=1

            if A[i][j] != res:
                print("NO")
                return
            else:
                res = 0

    print("YES")
    for i in B:
        for j in i:
            print(j, end=" ")
        print(end="\n")

OrMat()