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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.StringTokenizer;

public class OrInMatrix {
	public static void main(String[] args) {
		MyScanner sc = new MyScanner();
		
		int M = sc.nextInt();
		int N = sc.nextInt();
		
		int[][] A = new int[M][N];
		int[][] B = new int[M][N];
		
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < N; j++) {
				A[i][j] = 1;
				B[i][j] = sc.nextInt();
			}
		}
		
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < N; j++) {
				if (B[i][j] == 0) {
					for (int r = 0; r < M; r++) {
						A[r][j] = 0;
					}
					for (int c = 0; c < N; c++) {
						A[i][c] = 0;
					}
				}
			}
		}
		
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < N; j++) {
				if (B[i][j] == 1 && (rowSum(A, i) + colSum(A, j) == 0)) {
					System.out.println("NO");
					return;
				}
			}
		}
		
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < N; j++) {
				sb.append(A[i][j] + " ");
			}
			sb.append("\n");
		}
		
		System.out.println("YES");
		System.out.print(sb.toString());
	}
	
	public static int colSum(int[][] mat, int col) {
		int sum = 0;
		for (int r = 0; r < mat.length; r++) {
			sum += mat[r][col];
		}
		return sum;
	}
	
	public static int rowSum(int[][] mat, int row) {
		int sum = 0;
		for (int c = 0; c < mat[0].length; c++) {
			sum += mat[row][c];
		}
		return sum;
	}

	public static class MyScanner {
		BufferedReader br;
		StringTokenizer st;

		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		String nextLine() {
			String str = "";
			try { str = br.readLine(); }
			catch (IOException e) { e.printStackTrace(); }
			return str;
		}
	}
}