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 solve(A, n, m):
    rows = {}
    col = {}

    if n == 1 and m ==1:
        return True, A
    sm = False
    for i in range(n):
        s = 0
        for j in range(m):

            if A[i][j] == 1:
                sm = True
                s += 1
        rows[i] = s

    for i in range(m):
        s = 0
        for j in range(n):
            if A[j][i] == 1:
                s += 1
        col[i] = s


    B = [[0 for x in range(m)] for y in range(n)]
    flag = 0
    ones = 0
    for i in range(n):
        for j in range(m):
            if A[i][j] == 1:
                flag = 1
                if rows[i] == m and col[j] == n:
                    B[i][j] = 1

                    ones = 1
                elif rows[i] < m and col[j] < n:
                    return False, []

    if flag and ones:
        return True, B
    else:
        if not sm:
            return True, A
        return False, []
def printMat(Mat):
    for i in Mat:
        for j in i:
            print(j, end=' ')
        print()


N, M = map(int, input().split())
A = []

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

flag, mat = solve(A, N, M)

if flag:
    print('YES')
    printMat(mat)
else:
    print('NO')