/*
This solution uses mo's algorithm in 3d. Complexity is O(n^(5/3))
*/
#include "bits/stdc++.h"
using namespace std;
#define rep(i,a,b) for(int i=(a); i<(b); ++i)
#define all(x) x.begin(),x.end()
#define sz(x) int(x.size())
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<vi> vvi;
const int N = 100000;
const int B = pow(N,.666);
int cnt[N];
int active = 0;
int main(){
cin.tie(NULL),cin.sync_with_stdio(false);
int n,q; cin >> n >> q;
vector<array<int,2>> a(n);
rep(i,0,n) rep(j,0,2) cin >> a[i][j];
vvi ord(3,vi(n));
rep(i,0,3) iota(all(ord[i]),0);
rep(i,0,2) sort(all(ord[i]), [&](int l, int r){
return a[l][i] < a[r][i];
});
sort(all(ord[2]),[&](int l, int r){
return a[l][0] + a[l][1] < a[r][0] + a[r][1];
});
vvi vals(3,vi(n));
rep(i,0,2) rep(j,0,n) vals[i][j] = a[ord[i][j]][i];
rep(j,0,n) vals[2][j] = a[ord[2][j]][0] + a[ord[2][j]][1];
vector<array<int,3>> qs(q), s(q);
vi qi(q);
iota(all(qi),0);
rep(i,0,q){
rep(j,0,3) cin >> qs[i][j],
qs[i][j] = lower_bound(all(vals[j]),qs[i][j]) - begin(vals[j]);
int x = 0;
rep(j,0,3){
s[i][j] = qs[i][j]/B;
if (x) s[i][j] = n - s[i][j];
x ^= s[i][j]&1;
}
}
sort(all(qi),[&](int l, int r){
return s[l] < s[r];
});
array<int,3> cur = {n,n,n}; // none are active yet
vi sol(q);
for(auto id : qi){
rep(j,0,3){
while (cur[j] > qs[id][j]) {
cur[j]--;
cnt[ord[j][cur[j]]]++;
if (cnt[ord[j][cur[j]]] == 3) active++;
}while (cur[j] < qs[id][j]){
cnt[ord[j][cur[j]]]--;
if (cnt[ord[j][cur[j]]] == 2) active--;
cur[j]++;
}
}
sol[id] = active;
}
for(auto c : sol) cout << c << '\n';
}
