Dreamoon and Sets
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 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
1 1
5
1 2 3 5
2 2
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=;
bool f=;
char ch=' ';
while(!isdigit(ch))
{
f|=(ch=='-'); ch=getchar();
}
while(isdigit(ch))
{
s=(s<<)+(s<<)+(ch^); ch=getchar();
}
return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
if(x<)
{
putchar('-'); x=-x;
}
if(x<)
{
putchar(x+''); return;
}
write(x/);
putchar((x%)+'');
return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
int n,K;
int main()
{
int i;
R(n); R(K);
Wl((+(n-)*)*K);
for(i=;i<=n;i++)
{
int oo=((i-)*)*K;
W(K+oo); W(*K+oo); W(*K+oo); Wl(*K+oo);
}
return ;
}
/*
Input
1 1
Output
5
1 2 3 5 Input
2 2
Output
22
2 4 6 22
14 18 10 16
*/