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 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
inputCopy
1 1
outputCopy
5
1 2 3 5
inputCopy
2 2
outputCopy
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 .
给定N个集合，每个集合里面的数的 gcd ==k ，且每个数只可以属于一个集合，问最大的数最小应该是多少？
找一下规律：1，2，3，5；7，8，9，11；13，14，15，17…….，以6 为一个周期，这些都是满足互质的，那么gcd==k ，也就是*k即可，
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 200005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define pll pair<ll,ll>
ll quickpow(ll a, ll b) {
ll ans = 1;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a%b);
}
int main()
{
ios::sync_with_stdio(false);
int n, k;
cin >> n >> k;
int a = 1, b = 2, c = 3, d = 5;
cout << k * (d + 6 * (n - 1)) << endl;
for (int i = 0; i < n; i++) {
cout << a*k << ' ' << b*k << ' ' << c*k << ' ' << d*k << endl;
a += 6;
b += 6;
c += 6;
d += 6;
}
return 0;
}