#include <iostream>
#include <vector>
#include <array>
#include <map>
using namespace std;
using ll = long long;
int n, q;
vector <pair <ll, ll>> ppl;
vector <array <ll, 3>> queries;
struct stree {
#define lc v*2
#define rc v*2+1
vector <map <int, ll>> tree;
vector <vector <pair <ll, ll>>> tree_pfx;
stree (int s) {
tree.resize(s*4);
tree_pfx.resize(s*4);
}
void pushup(int v, int val) {
tree[v][val]++;
}
void update(int v, int tl, int tr, int p, ll val) {
if (tl > p || tr < p)
return;
if (tl == tr) {
tree[v][val]++;
return;
}
int tm = (tl + tr) / 2;
update(lc, tl, tm, p, val);
update(rc, tm+1, tr, p, val);
pushup(v, val);
}
void build_pfx(int v, int tl, int tr) {
int p = 0;
tree_pfx[v].resize(tree[v].size());
for (auto i : tree[v]) {
tree_pfx[v][p] = i;
if (p-1 >= 0)
tree_pfx[v][p].second += tree_pfx[v][p-1].second;
p++;
}
if (tl == tr)
return;
int tm = (tl + tr) / 2;
build_pfx(lc, tl, tm);
build_pfx(rc, tm+1, tr);
}
int query(int v, int tl, int tr, int l, int r, int val) {
if (tl > r || tr < l)
return 0;
if (tl >= l && tr <= r) {
pair <ll, ll> tp = {val, 0};
auto x = lower_bound(tree_pfx[v].begin(), tree_pfx[v].end(), tp) - tree_pfx[v].begin();
if (x == 0)
return tree_pfx[v].back().second;
else
return tree_pfx[v].back().second - tree_pfx[v][x-1].second;
}
int tm = (tl + tr) / 2;
int a = query(lc, tl, tm, l, r, val);
int b = query(rc, tm+1, tr, l, r, val);
return a + b;
}
};
void subtask1() {
for (int i = 0; i < q; i++) {
int ans = 0;
ll a = queries[i][0], b = queries[i][1], c = queries[i][2];
for (int j = 0; j < n; j++) {
if (ppl[j].first + ppl[j].second >= c && ppl[j].first >= a && ppl[j].second >= b)
ans++;
}
cout << ans << '\n';
}
}
void subtask2() {
int sz = 1e5 + 50;
stree tree(sz + 1);
for (int i = 0; i < n; i++) {
tree.update(1, 0, sz, ppl[i].first, ppl[i].second);
}
tree.build_pfx(1, 0, sz);
for (int i = 0; i < q; i++) {
cout << tree.query(1, 0, sz, queries[i][0], sz, queries[i][1]) << '\n';
}
}
int main() {
cin >> n >> q;
ppl.resize(n);
queries.resize(q);
for (int i = 0; i < n; i++)
cin >> ppl[i].first >> ppl[i].second;
for (int i = 0; i < q; i++)
cin >> queries[i][0] >> queries[i][1] >> queries[i][2];
if (n <= 3000 && q <= 3000)
subtask1();
else
subtask2();
}
