D. Hot Days

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The official capital and the cultural capital of Berland are connected by a single road running through?n?regions. Each region has a unique climate, so the?i-th?(1?≤?i?≤?n)?region has a stable temperature of?ti?degrees in summer.

This summer a group of?m?schoolchildren wants to get from the official capital to the cultural capital to visit museums and sights. The trip organizers transport the children between the cities in buses, but sometimes it is very hot. Specifically, if the bus is driving through the?i-th region and has?k?schoolchildren, then the temperature inside the bus is?ti? ?k?degrees.

Of course, nobody likes it when the bus is hot. So, when the bus drives through the?i-th region, if it has more than?Ti?degrees inside, each of the schoolchild in the bus demands compensation for the uncomfortable conditions. The compensation is as large as?xi?rubles and it is charged in each region where the temperature in the bus exceeds the limit.

To save money, the organizers of the trip may arbitrarily add or remove extra buses in the beginning of the trip, and between regions (of course, they need at least one bus to pass any region). The organizers can also arbitrarily sort the children into buses, however, each of buses in the?i-th region will cost the organizers?costi?rubles. Please note that sorting children into buses takes no money.

Your task is to find the minimum number of rubles, which the organizers will have to spend to transport all schoolchildren.

Input

The first input line contains two integers?n?and?m?(1?≤?n?≤?105;?1?≤?m?≤?106)?? the number of regions on the way and the number of schoolchildren in the group, correspondingly. Next?n?lines contain four integers each: the?i-th line contains?ti,?Ti,?xi?and?costi?(1?≤?ti,?Ti,?xi,?costi?≤?106). The numbers in the lines are separated by single spaces.

Output

Print the only integer ? the minimum number of roubles the organizers will have to spend to transport all schoolchildren.

Please, do not use the?%lld?specifier to read or write 64-bit integers in С . It is preferred to use cin, cout streams or the?%I64dspecifier.

Sample test(s)

input

2 1030 35 1 10020 35 10 10

output

120

input

3 10010 30 1000 15 10 1000 310 40 1000 100000

output

200065

Note

In the first sample the organizers will use only one bus to travel through the first region. However, the temperature in the bus will equal30? ?10?=?40?degrees and each of?10?schoolchildren will ask for compensation. Only one bus will transport the group through the second region too, but the temperature inside won't exceed the limit. Overall, the organizers will spend?100? ?10? ?10?=?120?rubles.

題意：n個(gè)區(qū)域，m個(gè)孩子坐車穿過這些區(qū)域，每個(gè)區(qū)域的溫度都不一樣，t，T，x，cost分別表示該地區(qū)的溫度t，車內(nèi)最大溫度限制T，車內(nèi)溫度若超過T對(duì)每個(gè)孩子的賠償x，坐車的費(fèi)用cost，求最小的費(fèi)用。

思路：1. 當(dāng) m <= (T-t) 時(shí)
最后的費(fèi)用都為 ： cost
2.m>(T-t) 時(shí)
有兩種選擇 一個(gè)是盡量做在一輛車上（對(duì)于補(bǔ)助費(fèi)比較少時(shí)）
sum=m*x;
另一個(gè)是 盡量多做車，但是要保證車都坐滿（剛好達(dá)到不要補(bǔ)助費(fèi)的時(shí)候）。?
sum=cost*( m%(T-t)==0? m/(T-t)+1: m/(T-t) );

代碼：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 1005#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;int main(){    __int64 n,m;    while (~scanf("%I64d%I64d",&n,&m))    {        __int64 t,T,x,c,ss1,ss2;        __int64 ans=0;        __int64 s1,s2;        for (__int64 i=0;i<n;i++)        {            s1=INF,s2;            scanf("%I64d%I64d%I64d%I64d",&t,&T,&x,&c);            if(T-t<=0)            {                ans+=m*x+c;                continue;            }            if (m<=T-t)            {                ans+=c;                continue;            }            ss1=c+m*x;            ss2=c*(m%(T-t)==0?m/(T-t):(m/(T-t)+1));            ans+=min(ss1,ss2);        }        printf("%I64d\n",ans);    }    return 0;}/*2 1030 35 1 10020 35 10 103 10010 30 1000 15 10 1000 310 40 1000 100000*/