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 isValidMatrix(matrix):
    for row in range(len(matrix)):
        for col in range(len(matrix[0])):
            if matrix[row][col] != 0:
                if not (rowOfOnes(matrix, row) or colOfOnes(matrix, col)):
                    return False
    return True

def rowOfOnes(matrix,row):
    for value in matrix[row]:
        if int(value) == 0:
            return False
    return True

def colOfOnes(matrix,col):
    for row in range(len(matrix)):
        if matrix[row][col] == 0:
            return False
    return True

def constructEmptyMatrix(rows,cols):
    matrix = []
    for row in range(rows):
        matrix += [[0] * cols]
    return matrix

def constructMatrixA(matrixA,matrixB):
    global rows
    global cols
    for row in range(rows):
        for col in range(cols):
            if (rowOfOnes(matrixB,row) and colOfOnes(matrixB,col)):
                matrixA[row][col] = 1
    return matrixA

def printMatrixA(matrixA):
    (rows,cols) = (len(matrixA),len(matrixA[0]))
    for row in range(rows):
        rowString = ""
        for col in range(cols):
            rowString += str(matrixA[row][col])
            rowString += " "
        rowString = rowString[:-1]
        print(rowString)

def hasNonZero(matrix):
    (rows,cols) = (len(matrixA),len(matrixA[0]))
    for row in range(rows):
        for col in range(cols):
            if matrix[row][col] != 0:
                return True
    return False

def validOutput(matrixA,matrixB):
    (rows,cols) = (len(matrixA),len(matrixA[0]))
    if hasNonZero(matrixA) and hasNonZero(matrixB):
        print("YES")
        printMatrixA(matrixA)
        return
    elif not hasNonZero(matrixA) and not hasNonZero(matrixB):
        print("YES")
        printMatrixA(matrixA)
        return
    print("NO")

start = input()
matrixDimensions = [int(x) for x in start.split(" ")]
rows = matrixDimensions[0]
cols = matrixDimensions[1]
matrixB = constructEmptyMatrix(rows,cols)
matrixA = constructEmptyMatrix(rows,cols)
for row in range(rows):
    rowString = input()
    rowValues = rowString.split(" ")
    for col in range(cols):
        matrixB[row][col] = int(rowValues[col])
if not isValidMatrix(matrixB):
    print("NO")
else:
    matrixA = constructMatrixA(matrixA,matrixB)
    validOutput(matrixA, matrixB)