可以离线，把询问拆成四个，然后把所有的按\(x\)坐标排序，这样就只要考虑\(y\)坐标了。然后把\(y\)坐标离散化，用树状数组统计即可

记得开longlong

//minamoto#include
    
     #define R register#define int long long#define fp(i,a,b) for(R int i=a,I=b+1;i
     
      I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)using namespace std;char buf[1<<21],*p1=buf,*p2=buf;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}int read(){    R int res,f=1;R char ch;    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');    return res*f;}char sr[1<<21],z[20];int C=-1,Z=0;inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}void print(R int x){    if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;    while(z[++Z]=x%10+48,x/=10);    while(sr[++C]=z[Z],--Z);sr[++C]='\n';}const int N=1e5+5;struct node{    int x,y,id,op,v;    node(){}    node(R int x,R int y,R int v,R int id=0,R int op=0):x(x),y(y),v(v),id(id),op(op){}    inline bool operator <(const node &b)const{return x==b.x?op