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<bits/stdc++.h>
#define R register
#define int long long
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(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]='
';
}
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<b.op:x<b.x;}
}q[N*5];int p[N*3],x[N],y[N],a[N],b[N],v[N],c[N],d[N],n,m,cc[N*3],lim,tot,ans[N];
inline void add(R int x,R int y){for(;x<=lim;x+=x&-x)cc[x]+=y;}
inline int query(R int x){R int res=0;for(;x;x-=x&-x)res+=cc[x];return res;}
signed main(){
// freopen("testdata.in","r",stdin);
n=read(),m=read();fp(i,1,n)x[i]=read(),p[++lim]=y[i]=read(),v[i]=read();
fp(i,1,m)a[i]=read(),p[++lim]=b[i]=read(),c[i]=read(),p[++lim]=d[i]=read();
sort(p+1,p+1+lim),lim=unique(p+1,p+1+lim)-p-1;
fp(i,1,n)y[i]=lower_bound(p+1,p+1+lim,y[i])-p,q[++tot]=node(x[i],y[i],v[i]);
fp(i,1,m){
b[i]=lower_bound(p+1,p+1+lim,b[i])-p,d[i]=lower_bound(p+1,p+1+lim,d[i])-p;
q[++tot]=node(a[i]-1,b[i]-1,1,i,1),q[++tot]=node(a[i]-1,d[i],-1,i,1);
q[++tot]=node(c[i],b[i]-1,-1,i,1),q[++tot]=node(c[i],d[i],1,i,1);
}sort(q+1,q+1+tot);
fp(i,1,tot)if(q[i].op==0)add(q[i].y,q[i].v);
else ans[q[i].id]+=query(q[i].y)*q[i].v;
fp(i,1,m)print(ans[i]);return Ot(),0;
}