//Sylwia Sapkowska
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef DEBUG
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
const int oo = 1e9+7;
struct UF{
vector<int>rep, sz;
vector<tuple<int, int, int, int>> st;
int s;
UF(int n){
s = n;
rep.assign(s, 0);
iota(rep.begin(), rep.end(), 0);
sz.assign(s, 1);
}
int f(int a){return a == rep[a] ? a : f(rep[a]);}
bool sameset(int a, int b){return f(a) == f(b);}
void undo(int t){
while ((int)st.size() > t){
auto [a, b, sA, sB] = st.back();
st.pop_back();
rep[a] = a;rep[b] = b;
sz[a] = sA;sz[b] = sB;
}
}
void u(int a, int b){
a = f(a); b = f(b);
if (a == b) return;
if (sz[a] < sz[b]) swap(a, b);
st.emplace_back(a, b, sz[a], sz[b]);
sz[a] += sz[b];
rep[b] = a;
}
};
void solve(){
int n, m, q; cin >> n >> m >> q;
int N = 2*n+2;
vector<pair<int, int>>edges;
for (int i = 0; i<m; i++){
int a, b; cin >> a >> b;
edges.emplace_back(a, b);
}
vector<int>R(m, oo);
UF dsu(N+2);
auto merge = [&](int i){
if (i == m) return true;
auto [a, b] = edges[i];
dsu.u(a, b+n);
dsu.u(a+n, b);
if (dsu.sameset(a, a+n)) return false;
return true;
};
function<void(int, int, int, int)>rec = [&](int l, int r, int a, int b){
//na przedziale [l, r] lewych końców odpowiedź należy do przedziału [a, b]
if (l > r) return;
if (a == b){
for (int i = l; i <= r; i++) R[i] = a;
return;
}
int mid = (l+r)/2;
bool ok = 1;
int t = (int)dsu.st.size();
for (int i = l; i < mid; i++) ok &= merge(i);
if (ok && R[mid] == oo){
for (int i = b; i >= mid; i--){ //odpowiedz in [a, b]
ok &= merge(i);
if (!ok){
R[mid] = i;
break;
}
}
if (R[mid] == oo) R[mid] = mid;
} else if (!ok) R[mid] = m;
dsu.undo(t);
// debug(l, r, mid, R[mid]);
//left side
ok = 1;
for (int i = b; i>R[mid]; i--) ok &= merge(i);
if (!ok) for (int i = l; i < mid; i++) R[i] = R[mid]; //is not bipartite
else rec(l, mid-1, a, R[mid]); //check next
dsu.undo(t);
//right side
ok = 1;
for (int i = l; i<=mid; i++) ok &= merge(i);
if (!ok) for (int i = mid+1; i<=r; i++) R[i] = m;
else rec(mid+1, r, R[mid], b);
dsu.undo(t);
};
rec(0, m-1, 0, m);
// debug(R);
while (q--){
int a, b; cin >> a >> b;
--a;--b;
cout << ((R[a] <= b) ? "NO\n" : "YES\n");
}
}
int32_t main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}
