Problem: Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column. 

We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct. 

The magnitude of a matrix equals to the maximum of b_k.

For example, suppose we have the following matrix:

\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix} 

We construct the array b:

  1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1; 
  2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4; 
  3. b_3 is the greatest common divisor of 2 and 4, that is 2; 
  4. b_4 is the greatest common divisor of 9 and 144, that is 9; 
  5. b_5 is the greatest common divisor of 7 and 84, that is 7. 



So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.

For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0. 

Input

The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.

Output

If there is no solution, output a single integer 0.

Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.

Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).

Examples

Input


2 2


Output


4 12
2 9

Input


1 1


Output


0

Note

In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.

In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main {
	static final int MOD_PRIME = 1000000007;

	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		Task solver = new Task();

		int i = 0;
		int t = 1;
		// t = in.nextInt();
		for (; i < t; i++)
			solver.solve(i, in, out);

		out.close();
	}

	static class Task {
		public void solve(int testNumber, InputReader in, PrintWriter out) {
			int r = in.nextInt();
			int c = in.nextInt();
			int[][] ans = new int[r][c];
			if (Math.min(r, c) == 1) {
				if (Math.max(r, c) == 1) {
					out.println(0);
					return;
				} else {
					int ctr = 2;
					for (int i = 0; i < r; i++) {
						for (int j = 0; j < c; j++) {
							ans[i][j] = ctr;
							ctr++;
						}
					}
				}
			} else {
				for (int i = 0; i < r; i++) {
					for (int j = 0; j < c; j++) {
						ans[i][j] = (i + 1) * (j + 1 + r);
					}
				}
			}
			for(int i = 0; i < r; i++) {
				for(int j = 0; j < c; j++) {
					out.print(ans[i][j] + " ");
				}
				out.println();
			}

		}
	}

	// template code

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

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

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

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

	}

	static long modexp(long a, long b, long p) {
		// returns a to the power b mod p by modular exponentiation

		long res = 1;
		long mult = a % p;
		while (b > 0) {
			if (b % 2 == 1) {
				res = (res * mult) % p;
			}
			b /= 2;
			mult = (mult * mult) % p;
		}

		return res;
	}

	static double log(double arg, double base) {
		// returns log of a base b, contains floating point errors, dont use for exact
		// calculations.

		if (base < 0 || base == 1) {
			throw new ArithmeticException("base cant be 1 or negative");
		}
		if (arg < 0) {
			throw new ArithmeticException("log of negative number undefined");
		}
		return Math.log10(arg) / Math.log10(base);
	}

	static int gcd(int a, int b) {
		if (b == 0) {
			return a;
		}

		return gcd(b, a % b);

	}

	static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		}

		return gcd(b, a % b);

	}

	// scope for improvement
	static ArrayList<Integer> sieveOfEratosthenes(int n) {
		boolean[] check = new boolean[n + 1];
		ArrayList<Integer> prime = new ArrayList<Integer>();
		for (long i = 2; i <= n; i++) {
			if (!check[(int) i]) {
				prime.add((int) i);
				for (long j = i * i; j <= n; j += i) {
					check[(int) j] = true;
				}
			}
		}

		return prime;
	}

	static int modInverse(int a, int n) {
		// returns inverse of a mod n by extended euclidean algorithm
		int t = 0;
		int newt = 1;
		int r = n;
		int newr = a;
		int quotient;
		int tempr, tempt;
		while (newr != 0) {
			quotient = r / newr;
			tempt = newt;
			tempr = newr;
			newr = r - quotient * tempr;
			newt = t - quotient * tempt;
			t = tempt;
			r = tempr;
		}
		if (r > 1) {
			throw new ArithmeticException("inverse of " + a + " mod " + n + " does not exist");
		} else {
			if (t < 0) {
				t += n;
			}
			return t;
		}
	}

	static int primeModInverse(int a, int n) {
		// returns inverse of a mod n by mod exponentiation, use only if n is prime
		return (int) modexp(a, n - 2, n);
	}
}