#include <bits/stdc++.h>
using namespace std;
#define int long long
#define FOR(I, L, R) for(int I(L) ; I <= (int)R ; ++I)
#define FOD(I, R, L) for(int I(R) ; I >= (int)L ; --I)
#define FOA(I, A) for(auto &I : A)
#define print(A,L,R) FOR(OK, L, R){if(A[OK]<=-oo / 10||A[OK]>=oo)cout<<"- ";else cout<<A[OK]<<' ';}cout<<'\n';
#define prints(A) FOA(OK, A){cout<<OK<<' ';}cout << '\n';
#define printz(A,L,R) FOR(OK, 0, L){FOR(KO, 0, R){if(A[OK][KO]>-oo&&A[OK][KO]<oo)cout<<A[OK][KO]<<' ';else cout << "- ";} cout << '\n';}cout << '\n';
#define fs first
#define sd second
#define ii pair<int,int>
#define iii pair<int, ii>
#define all(A) A.begin(), A.end()
#define quickly cin.tie(0) -> ios_base::sync_with_stdio(0);
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int oo = 1e18;
struct query{
int x, y, z, in;
} Q[N];
int n, q;
ii a[N];
struct Segment_tree{
int sg[N << 2];
void update(int id, int l, int r, int pos, int val){
if(l > pos || r < pos){
return;
}
if(l == r){
sg[id] += val;
return;
}
int mid = (l + r) >> 1;
update(id << 1, l, mid, pos, val);
update(id << 1 | 1, mid + 1, r, pos, val);
sg[id] = sg[id << 1] + sg[id << 1 | 1];
}
int get(int id, int l, int r, int u, int v){
if(l > v || r < u){
return 0;
}
if(u <= l && r <= v){
return sg[id];
}
int mid = (l + r) >> 1;
return get(id << 1, l, mid, u, v) + get(id << 1 | 1, mid + 1, r, u, v);
}
};
namespace Subtask1{
void solve(){
while(q--){
int ans = 0;
int x, y, z;
cin >> x >> y >> z;
FOR(i, 1, n){
if(a[i].fs >= x && a[i].sd >= y && a[i].fs + a[i].sd >= z){
ans++;
}
}
cout << ans << '\n';
}
}
}
namespace Subtask2{
static Segment_tree sg;
static int ans[N];
void solve(){
sort(a + 1, a + 1 + n, greater<ii>());
FOR(i, 1, q){
int x, y, z;
cin >> x >> y >> z;
Q[i] = {x, y, z, i};
}
sort(Q + 1, Q + 1 + q, [&](query &u, query &v){
return u.x > v.x;
});
int en = N - 5;
int l = 1;
FOR(i, 1, q){
while(l <= n && a[l].fs >= Q[i].x){
sg.update(1, 1, en, a[l].sd, 1);
++l;
}
ans[Q[i].in] = sg.get(1, 1, en, Q[i].y, en);
}
FOR(i, 1, q){
cout << ans[i] << '\n';
}
}
}
signed main(){ quickly
cin >> n >> q;
FOR(i, 1, n){
cin >> a[i].fs >> a[i].sd;
}
if(n <= 3000){
Subtask1::solve();
}
else{
Subtask2::solve();
}
}
