Petya and Origami

Posted 2021-01-22 staceyacm

Petya is having a party soon, and he has decided to invite his nn friends.

He wants to make invitations in the form of origami. For each invitation, he needs two red sheets, five green sheets, and eight blue sheets. The store sells an infinite number of notebooks of each color, but each notebook consists of only one color with kk sheets. That is, each notebook contains kk sheets of either red, green, or blue.

Find the minimum number of notebooks that Petya needs to buy to invite all nn of his friends.

Input

The first line contains two integers nn and kk (1≤n,k≤1081≤n,k≤108) — the number of Petya‘s friends and the number of sheets in each notebook respectively.

Output

Print one number — the minimum number of notebooks that Petya needs to buy.

Petya is having a party soon, and he has decided to invite his nn friends.

He wants to make invitations in the form of origami. For each invitation, he needs two red sheets, five green sheets, and eight blue sheets. The store sells an infinite number of notebooks of each color, but each notebook consists of only one color with kk sheets. That is, each notebook contains kk sheets of either red, green, or blue.

Find the minimum number of notebooks that Petya needs to buy to invite all nn of his friends.

Input

The first line contains two integers nn and kk (1≤n,k≤1081≤n,k≤108) — the number of Petya‘s friends and the number of sheets in each notebook respectively.

Output

Print one number — the minimum number of notebooks that Petya needs to buy.

Examples

input

Copy

3 5
output

Copy

10
input

Copy

15 6
output

Copy

38
Note

In the first example, we need 22 red notebooks, 33 green notebooks, and 55 blue notebooks.

In the second example, we need 55 red notebooks, 1313 green notebooks, and 2020 blue notebooks.

题解：此题太水，基本都会

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
 
int main()
{
    int n,k;
    cin>>n>>k;
    int red=n*2/k;
    int green=n*5/k;
    int blue=n*8/k;
    if(red<n*2.0/k)
    {
        red=red+1;
    }
    if(green<n*5.0/k)
    {
        green=green+1;
    }
    if(blue<n*8.0/k)
    {
        blue+=1;
    }
    cout<<red+green+blue<<endl;
    return 0;
 }