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 JAVA):

Solution:
import java.util.Scanner;

public class B_OrInMatrix {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n, m;
        m = sc.nextInt(); n = sc.nextInt();
        boolean B[][] = new boolean[m][n], A[][] = new boolean[m][n], C[][] = new boolean[m][n];
        for (int i = 0; i < m; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                B[i][j] = sc.nextInt()==1;
                A[i][j] = true;
                C[i][j] = false;
            }
        }

        for (int i = 0; i < m; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                if (!B[i][j])
                {
                    for (int k = 0; k < n; ++k)
                    {
                        A[i][k] = false;
                    }
                    for (int k = 0; k < m; ++k)
                    {
                        A[k][j] = false;
                    }

                }
            }
        }

        for (int i = 0; i < m; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                if (A[i][j] )
                {
                    for (int k = 0; k < n; ++k)
                    {
                        C[i][k] = true;
                    }
                    for (int k = 0; k < m; ++k)
                    {
                        C[k][j] = true;
                    }

                }
            }
        }

        for (int i = 0; i < m; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                if (C[i][j] != B[i][j])
                {
                    System.out.println("NO");
                    return;
                }
            }
        }

        System.out.println("YES");
        for (int i = 0; i < m; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                System.out.print((A[i][j]?1:0) + " ");
            }
            System.out.println();
        }


        return;
    }
}