#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define FOR(i, l, r) for(int i = l; i <= r; i++)
#define FORD(i, l, r) for(int i = l; i >= r; i--)
#define db double
#define ldb long double
#define all_1(x) (x).begin() + 1, (x).end()
#define all(x) (x).begin(), (x).end()
#define ins insert
#define pb push_back
int n, m, q;
struct segment_tree{
int _n;
vector<int> _st;
segment_tree(int n = 0){
init(n);
}
void init(int n){
_n = n;
_st.assign(4 * _n + 5, -2e9);
}
void update(int i, int l, int r, int pos, int val){
if(pos > r || pos < l) return;
if(l == r){
_st[i] = max(_st[i], val);
return;
}
int m = (l + r) >> 1, cl = (i << 1), cr = (cl | 1);
update(cl, l, m, pos, val);
update(cr, m + 1, r, pos, val);
_st[i] = max(_st[cl], _st[cr]);
}
int query(int i, int l, int r, int u, int v){
if(l > v || r < u) return -2e9;
if(l >= u && r <= v) return _st[i];
int m = (l + r) >> 1, cl = (i << 1), cr = (cl | 1);
return max(query(cl, l, m, u, v), query(cr, m + 1, r, u, v));
}
};
struct Data{
int from, to, id;
};
void solve(){
cin >> n >> m >> q;
vector<pair<int, int>> curtain(m + 1);
FOR(i, 1, m) cin >> curtain[i].first >> curtain[i].second;
vector<Data> queries(q + 1);
FOR(i, 1, q) cin >> queries[i].from >> queries[i].to, queries[i].id = i;
sort(all_1(curtain), [&](pair<int, int> a, pair<int, int> b){
return a.second < b.second || (a.second == b.second && a.first < b.first);
});
sort(all_1(queries), [&](Data a, Data b){
return a.to < b.to || (a.to == b.to && a.from < b.from);
});
segment_tree st(n);
vector<bool>ans(q + 1);
int j = 1;
FOR(qr, 1, q){
auto [from, to, id] = queries[qr];
while(j <= m && curtain[j].second <= to){
st.update(1, 1, n, curtain[j].first, curtain[j].second);
++j;
}
int i = from;
bool ok = true;
while(i <= to){
int maxi = st.query(1, 1, n, from, i);
if(maxi < i){
ok = false;
break;
}
i = maxi + 1;
}
ans[id] = ok;
}
FOR(i, 1, q) cout << (ans[i] ? "YES\n" : "NO\n");
}
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
int t = 1;
//cin >> t;
while(t--){
solve();
}
return 0;
}
/*
n san khau, m man
man i che l[i] -> r[i]
truy van j, yeu can che doan from[j] -> to[j]
from[j] -> to[j]
l[i] < from[j] || r[i] > to[j] -> continue
=> l[i] >= from[j] && r[i] <= to[j] -> thang nao cung dc
gia su ta co dinh dc to[j]
chi quan tam max cua r[j],
*/
