Algorithm
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.
The single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).
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.
1 1
5
1 2 3 5
2 2
22
2 4 6 22
14 18 10 16
For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since 
.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
int n, k, mx;
vector<vector<int> > sol;
int main() {
scanf("%d %d", &n, &k);
sol.resize(n);
for(int i = 0; i < n; ++i) {
sol[i].push_back(6 * i + 1);
sol[i].push_back(6 * i + 2);
sol[i].push_back(6 * i + 3);
sol[i].push_back(6 * i + 5);
}
for(int i = 0; i < sol.size(); ++i)
for(int j = 0; j < 4; ++j)
mx = max(mx, sol[i][j] * k);
printf("%d\n", mx);
for(int i = 0; i < sol.size(); ++i, puts(""))
for(int j = 0; j < 4; ++j)
printf("%s%d", j == 0 ? "" : " ", sol[i][j] * k);
return 0;
}Input
Output
1 2 3 5
Demonstration
Codeforces Solution-Dreamoon and Sets-Solution in C, C++, Java, Python