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.awt.*;
import java.util.ArrayList;
import java.util.Scanner;

public class ORinMatrix {
    static int[][]matrix,newMatrix;
    static int n,m,cnt;
    static int[][] fillRowZeros(int row,int from){
        for (int i = 0;i<m;i++){
                newMatrix[row][i] = 2;
        }
        return newMatrix;
    }
    static int[][] fillColZeros(int col,int from){
        for (int i = 0;i<n;i++){
            newMatrix[i][col]= 2;
        }
        return newMatrix;
    }
    static boolean isItPossibleRow(int row,int from){
        for (int i = 0;i<m;i++){
            if(newMatrix[row][i]!=2){
                newMatrix[row][i] = 1;
                return true;
            }
        }
        return false;
    }
    static boolean isItPossibleCol(int col,int from){
        for (int i = 0;i<n;i++){
            if(newMatrix[i][col]!=2){
                newMatrix[i][col] = 1;
                return true;
            }
        }
        return false;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
         n = sc.nextInt();
         m = sc.nextInt();
         matrix = new int[n][m];
         newMatrix = new int[n][m];
        ArrayList<Point> ones = new ArrayList<>();
        for(int i = 0;i<n;i++){
            for (int j = 0; j< m;j++){
                matrix[i][j] = sc.nextInt();
                if(matrix[i][j]==0){
                    newMatrix = fillColZeros(j,i);
                    newMatrix =  fillRowZeros(i,j);
                }else{
                    ones.add(new Point(i,j));
                }
            }
        }

       for (int i = 0;i<ones.size();i++){
           Point temp = ones.get(i);
           if(!isItPossibleRow(temp.x,temp.y) && !isItPossibleCol(temp.y,temp.x)){
               System.out.println("NO");
               return;
           }
       }
        System.out.println("YES");
       for (int i=0;i<n;i++){
           for (int j = 0;j<m;j++){
               System.out.print((newMatrix[i][j]==2?0:1) + " ");
           }
           System.out.println();
       }

    }
}