Problem: Dreamoon likes to play with sets, integers and <image>. <image> is defined as the largest positive integer that divides both a and b.

Let S be a set of exactly four distinct integers greater than 0. Define S to be of rank k if and only if for all pairs of distinct elements si, sj from S, <image>.

Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print one possible solution.

Input

The single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).

Output

On the first line print a single integer — the minimal possible m. 

On each of the next n lines print four space separated integers representing the i-th set.

Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one of them.

Examples

Input

1 1


Output

5
1 2 3 5


Input

2 2


Output

22
2 4 6 22
14 18 10 16

Note

For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since <image>.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class DreamoonAndSets {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        
        int N = sc.nextInt();
        long K = sc.nextLong();
        
        StringBuilder sb = new StringBuilder();
        long max = 5 * K;
        for (int i = 0; i < N; i++) {
        	int curr = 6 * i;
        	int x1 = curr + 1;
        	int x2 = curr + 2;
        	int x3 = curr + 3;
        	int x4 = curr + 5;
        	sb.append((K * x1) + " " + (K * x2) + " " + (K * x3) + " " + (K * x4) + "\n");
        	max = K * x4;
        }
        
        System.out.println(max);
        System.out.print(sb.toString());
    }
    
    public static int binSearchBig(Integer[] arr, Integer val) {
    	int idx = Arrays.binarySearch(arr, val);
    	if (idx >= 0) {
    		return arr[idx];
    	} else {
    		return arr[-idx - 1];
    	}
    }

    /**
     * Computes all the primes up to the specified number. 
     */
    public static Integer[] primesTo(int n)
    {
        ArrayList<Integer> primes = new ArrayList<Integer>();

        boolean[] prime = new boolean[n + 1];
        for (int i = 0; i < prime.length; i++)
            prime[i] = true;

        prime[0] = false;
        prime[1] = false;
        for (int i = 2; i < prime.length; i++)
        {
            if (prime[i])
            {
                primes.add(i);
                for (int j = 2 * i; j < prime.length; j += i)
                {
                    prime[j] = false;
                }
            }
        }

        return primes.toArray(new Integer[0]);
    }

    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;
        }
    }
}