Petya and Origami

最新推荐文章于 2022-01-07 11:22:00 发布

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最新推荐文章于 2022-01-07 11:22:00 发布

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分类专栏： codefores水题

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本文探讨了如何为派对制作折纸邀请函，计算为了邀请n位朋友，需要购买的最少笔记本数量，每个笔记本包含k张特定颜色的纸张，用于制作邀请函所需的红、绿、蓝三种颜色的纸张。

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.

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

input

Copy

15 6

output

Copy

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;
 }