#include <bits/stdc++.h>
using namespace std;
#define int long long
// the answer for a set of stores and some point x is the minimum integer m such that MEX([x - m...x + m]) = k + 1
// MEX(l..r) = min(MEX(1..r), MEX(l..n))
const int MX = 1e8+1;
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n, k, q; cin >> n >> k >> q;
vector<vector<int>> vals(MX);
vector<int> pref(MX), suff(MX);
for (int i = 1; i <= n; i++) {
int x, t, a, b; cin >> x >> t >> a >> b;
vals[x].push_back(t);
}
int currMex = 0;
vector<bool> ex(k+1, false);
for (int i = 1; i < MX; i++) {
for (auto v: vals[i]) ex[v] = true;
while (ex[currMex]) currMex++;
pref[i] = currMex;
}
fill(ex.begin(), ex.end(), false);
currMex = 0;
for (int i = MX-1; i >= 1; i--) {
for (auto v: vals[i]) ex[v] = true;
while (ex[currMex]) currMex++;
suff[i] = currMex;
}
for (int i = 1; i <= q; i++) {
int x, y; cin >> x >> y;
int lo = 0, hi = MX;
int ans = -1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
int ll = max(1LL, x - mid);
int rr = min(MX, x + mid);
int val = min(pref[rr], suff[ll]);
if (val == k+1) {
ans = mid;
hi = mid-1;
}
else lo = mid+1;
}
cout << ans << "\n";
}
}
