codeforces476D

Dreamoon and Sets

 CodeForces - 476D 

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 s_i, s_j from S, .

Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to msuch 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 .

sol：构造出来的肯定是四个互质的数字*K，对于这四个互质的数字要尽量小，容易构造出这样四个

1 2 3 5

7 8 9 11   （每次加6）

没有比上面更小的了

/*
输出n组集合，每组4个。对于任意一组中的4个元素，一组内任意2个数的gcd==k
且n组的所有数字各不相同。要使得n组中最大的数字最小。
*/
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
int n,K;
int main()
{
    int i;
    R(n); R(K);
    Wl((5+(n-1)*6)*K);
    for(i=1;i<=n;i++)
    {
        int oo=((i-1)*6)*K;
        W(K+oo); W(2*K+oo); W(3*K+oo); Wl(5*K+oo);
    }
    return 0;
}
/*
Input
1 1
Output
5
1 2 3 5

Input
2 2
Output
22
2 4 6 22
14 18 10 16
*/