D. Dreamoon and Sets（Codeforces Round #273)，

D. Dreamoon and Sets（Codeforces Round #273)，
D. Dreamoon and Setstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Dreamoon likes to play with sets, integers and .  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 fromS, .

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.

Sample test(s)input

1 1

output

51 2 3 5

input

2 2

output

222 4 6 2214 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 .

構造，當k等於1時，推幾組資料，例如1,2,3,5；7,8,9,11；13,14,15,17；19,20,21,23；25,26,27,29。就會發現是以6為周期，而對每個周期內的數乘以k就會使周期內的數兩兩的最大公約數為k。

代碼：

#include <cstdio>#include <iostream>#include <cstring>using namespace std;int main(){    int n, k;    scanf("%d %d", &n, &k);    int a = 1, b = 2, c = 3, d = 5;    printf("%d\n", (d * k + 6 * k * (n- 1)));    a*=k;    b*=k;    c*=k;    d*=k;    for(int i = 0; i < n; i++)    {        printf("%d %d %d %d\n",a, b, c, d);        a += 6 * k;        b += 6 * k;        c += 6 * k;        d += 6 * k;    }}

codeforces怎才可以改變id顏色（程式設計的）

不是北京時間,晚4個小時這樣吧.顏色的話不是很清楚...我只知道兩三百名就是藍名