/*
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
*/
#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
using namespace std;
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int>vi;
typedef vector<vector<int>>vvi;
typedef vector<ll>vl;
typedef vector<vl> vvl;
typedef pair<int,int>pi;
typedef pair<ll,ll> pl;
typedef vector<pl> vpl;
typedef vector<ld> vld;
typedef pair<ld,ld> pld;
typedef vector<pi> vpi;
//typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
template<typename T> ostream& operator<<(ostream& os, vector<T>& a){os<<"[";for(int i=0; i<ll(a.size()); i++){os << a[i] << ((i!=ll(a.size()-1)?" ":""));}os << "]\n"; return os;}
#define all(x) x.begin(),x.end()
#define YES out("YES")
#define NO out("NO")
#define out(x){cout << x << "\n"; return;}
#define outfl(x){cout << x << endl;return;}
#define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define print(x){for(auto ait:x) cout << ait << " "; cout << "\n";}
#define pb push_back
#define umap unordered_map
template<typename T>
void read(vector<T>& v){
int n=v.size();
for(int i=0; i<n; i++)
cin >> v[i];
}
template<typename T>
vector<T>UNQ(vector<T>a){
vector<T>ans;
for(T t:a)
if(ans.empty() || t!=ans.back())
ans.push_back(t);
return ans;
}
void solve();
int main(){
GLHF;
int t=1;
//cin >> t;
while(t--)
solve();
}
struct seg{
seg* lp=0,*rp=0;
int l,r,m;
int sum=0;
seg(int l,int r):l(l),r(r),m((l+r)/2){}
void expand(){
if(!lp)lp=new seg(l,m);
if(!rp)rp=new seg(m,r);
}
void upd(int i,int v){
sum+=v;
if(l+1==r)
return;
expand();
if(i<m)
lp->upd(i,v);
else
rp->upd(i,v);
}
int qur(int a,int b){
if(b<=l || r<=a)
return 0;
if(a<=l && r<=b)
return sum;
expand();
return lp->qur(a,b) + rp->qur(a,b);
}
};
/*
Basically what we do is split the queries into 2 types:
1. queries where A+B>=C, then we only need to count S>=A, T>=B, (and then it applies that S+T>=A+B>=C).
So we only need to count how many fulfil S>=A, T>=B.
This can be done offline - sorting queries and pairs by S/A, then querying a segment tree on [B,inf].
*/
void solve() {
int n,q;
cin >> n >> q;
vvl a(n,vl(2));
for(int i=0 ;i<n; i++)
cin >> a[i][0] >> a[i][1];
vvl queries(q,vl(4));
for(int i=0; i<q; i++){
cin >>queries[i][0] >> queries[i][1] >> queries[i][2];
queries[i][3]=i;
}
sort(all(a));
sort(all(queries));
vi ans(q);
seg ST(0,1e9 +10);
for(int i=0; i<n; i++)
ST.upd(a[i][1],1);
for(int i=0,j=0; i<q; i++){
while(j<n && a[j][0]<queries[i][0])
ST.upd(a[j++][1],-1);
if(queries[i][0]+queries[i][1]>=queries[i][2])
ans[queries[i][3]] = ST.qur(queries[i][1],1e9 + 2);
}
sort(all(a),[&](vl v1,vl v2){
return v1[0]+v1[1] < v2[0]+v2[1];
});
sort(all(queries),[&](vl v1,vl v2){
return v1[2]<v2[2];
});
ST = seg(0,1e9 + 10);
for(int i=0; i<n;i++)
ST.upd(a[i][1],1);
for(int i=0,j=0; i<q; i++){
ll A=queries[i][0],B=queries[i][1],C=queries[i][2],ix=queries[i][3];
if(A+B>=C)
continue;
while(j<n && a[j][0]+a[j][1]<C)
ST.upd(a[j][1],-1), j++;
ans[ix] = ST.qur(B,1e9 + 2);
}
ST = seg(0,1e9 + 10);
for(int i=0; i<n;i++)
ST.upd(a[i][0],1);
for(int i=0,j=0; i<q; i++){
ll A=queries[i][0],B=queries[i][1],C=queries[i][2],ix=queries[i][3];
if(A+B>=C)
continue;
while(j<n && a[j][0]+a[j][1]<C)
ST.upd(a[j][0],-1), j++;
ans[ix] -= ST.qur(0,A);
}
for(int x:ans)
cout << x << "\n";
}
