思路:需要手玩一会....



#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    printf("%d\n",(n-1)*k*6+5*k);
    for(int i=0;i<n;i++)
        printf("%d %d %d %d\n",i*k*6+k,i*k*6+2*k,i*k*6+3*k,i*k*6+5*k);
}




D. Dreamoon and Sets



time limit per test



memory limit per test



input



output


Dreamoon likes to play with sets, integers and 

Codeforces Round #272 (Div. 2) D Dreamoon and Sets(构造)_i++


Codeforces Round #272 (Div. 2) D Dreamoon and Sets(构造)_sed_02

 is defined as the largest positive integer that divides both a and b.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 sisjfrom S

Codeforces Round #272 (Div. 2) D Dreamoon and Sets(构造)_sed_03

.

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


Input


nk (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).


Output


m.

n lines print four space separated integers representing the i-th set.

m, print any one of them.


Examples


input


1 1


output


51 2 3 5


input


2 2


output


222 4 6 22 14 18 10 16


Note

{1, 2, 3, 4} isn't a valid set of rank 1 since 

Codeforces Round #272 (Div. 2) D Dreamoon and Sets(构造)_sed_04

.