#include<bits/stdc++.h>
using namespace std;
#include<ext/pb_ds/tree_policy.hpp>
#include<ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#define ordered_set tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>
#define ll long long
#define pii pair<ll,ll>
#define mp make_pair
#define f first
#define s second
#define FOR(i,j,n,k) for(int i = j;((k>0)?(i<n):(i>n));i+=k)
#define pb push_back
#define all(x) x.begin(),x.end()
#define siz(x) (int)sizeof(x)
ll n, m, k, q, l, r, x, y, z;
const ll tas = 5e5 + 10;
ll a[tas];
vector<int> grid[tas];
string s;
ll ans[tas];
ll t[4 * tas];
void push(int i, int l, int r) {
if (t[2 * i + 1] < t[i]) {
t[2 * i + 1] = t[i];
}
if (t[2 * i + 2] < t[i]) {
t[2 * i + 2] = t[i];
}
}
ll check(ll x, ll y) {
if (x == -1)return y;
if (y == -1) return x;
return min(x, y);
}
void update(int i, int l, int r, int tl, int tr, ll val) {
if (r < tl || tr < l) {
return;
}
if (tl <= l && r <= tr) {
t[i] = max(t[i], val);
// cout << i << " " << t[i] << endl;
return;
}
push(i, l, r);
int mid = (l + r) / 2;
update(2 * i + 1, l, mid, tl, tr, val);
update(2 * i + 2, mid + 1, r, tl, tr, val);
t[i] = check(t[2 * i + 1], t[2 * i + 2]);
}
ll query(int i, int l, int r, int tl, int tr) {
// cout << r << endl;
if (r < tl || tr < l) {
return 1e12;
}
if (tl <= l && r <= tr) {
// cout << l << " " << t[i] << endl;
return t[i];
}
push(i, l, r);
int mid = (l + r) / 2;
return min(query(2 * i + 1, l, mid, tl, tr),
query(2 * i + 2, mid + 1, r, tl, tr));
}
void solve(int tc = 0) {
cin >> n >> m >> q;
FOR(i, 0, 4 * n, 1)t[i] = -1e12;
vector<pair<int, pii>> vec;
FOR(i, 0, m, 1) {
cin >> l >> r;
l--, r--;
vec.pb(mp(r, mp(0, l)));
}
FOR(i, 0, q, 1) {
cin >> l >> r;
l--, r--;
// cout << i + 1 << endl;
vec.pb(mp(r, mp(i + 1, l)));
}
sort(all(vec));
for (auto val : vec) {
l = val.s.s, r = val.f;
// cout << val.s.f << endl;
if (val.s.f == 0) {
update(0, 0, n - 1, l, r, l);
}
else {
// cout << query(0, 0, n - 1, l, r) << " " << l << endl;
ans[val.s.f - 1] = (query(0, 0, n - 1, l, r) >= l && 1e10 > query(0, 0, n - 1, l, r));
}
}
// FOR(i, 0, n, 1) cout << i << " " << query(0, 0, n - 1, i, i) << endl;
FOR(i, 0, q, 1) if (ans[i]) cout << "YES" << endl;
else cout << "NO" << endl;
}
int main() {
int tc = 1;
// cin >> tc;
FOR(i, 0, tc, 1) solve(i);
}
