The road of calculation (knockout car)

The racing race has become more and more popular in Pandora. But their game is not the same as our usual: N-Cars compete on a long straight track. The speed of each car is 1m/s, and the entire track is marked with coordinates at every metre.

In racing cars, I moved from ai to bI from 0 seconds onwards. After reaching Bi , I move back from BI to ai . Cycle.

Another garlic fungus! Originally this is the garlic fungus is playing a mobile phone game. Garlic fungus can put down TNT in some places to blow up some cars. Because he has a m problem. The question J is: how many cars are there in the TJ moment between xi and yi ?

Your task is to answer the question of Meng garlic.

Input

The first line of input contains two numbers n and M (1≤n, m≤1000), respectively, representing the number of racing cars and the number of garlic problems in the race.

i , Bi (0≤ai , Bi  ≤109 , Ai  != bi ", respectively, representing the starting and ending points of the racing I.

j , Yj , Tj (0≤xj  ≤yj  ≤109 , 0≤tj  ≤109 ", respectively, representing the left and right coordinate boundaries of question J and the time of inquiry.

Output

The output has a total of m rows, each with an integer representing the answer to the corresponding m question.

Example 1

Input:

5 50 10 22 33 54 50 5 00 1 20 2 12 5 22 5 3

Output:

51243

#include <stdio.h>const int N = 1005;struct car{int l,r,t,flag;}    Node[n];int Main () {int n,m,ans,l,r,t; while (scanf ("%d%d", &n,&m) >0) {for (int i=0; i<n; i++) {scanf ("%d%d", &node[i].l,&nod            E[I].R);            if (NODE[I].L&LT;=NODE[I].R) node[i].flag=0;                else{node[i].flag=1;                int tmp=node[i].l;                NODE[I].L=NODE[I].R;            node[i].r=tmp;        } NODE[I].T=NODE[I].R-NODE[I].L;            } while (m--) {scanf ("%d%d%d", &l,&r,&t);            ans=0;                for (int i=0; i<n; i++) {int tmp=t/node[i].t+node[i].flag;                int res=t%node[i].t;                int Loc;                if (tmp&1) loc=node[i].r-res;                else Loc=node[i].l+res;            if (loc>=l&&loc<=r) ans++; } printf ("%d\n", ans); }    }}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

The road of calculation (knockout car)