// not by me :P
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main(){
ios::sync_with_stdio(false);
cin.tie(NULL);
int N, M, K;
if(!(cin >> N >> M >> K)) return 0;
vector<vector<ll>> costs(M + 1);
for(int i = 0; i < N; i++){
int a;
ll c;
cin >> a >> c;
costs[a].push_back(c);
}
vector<int> x(M + 1), y(M + 1);
for(int j = 1; j <= M; j++){
cin >> x[j] >> y[j];
}
// 1) sort and prefix sums
vector<ll> base_pref(M + 1, 0);
ll base_cost = 0;
vector<vector<ll>> extra(M + 1);
for(int j = 1; j <= M; j++){
auto &v = costs[j];
sort(v.begin(), v.end());
int need = x[j];
if((int)v.size() < need){
// not enough items -> no feasible plans
for(int i = 0; i < K; i++) cout << -1 << '\n';
return 0;
}
// compute prefix
vector<ll> pref(v.size() + 1, 0);
for(size_t i = 1; i < pref.size(); i++) pref[i] = pref[i-1] + v[i-1];
base_cost += pref[need];
// build extra costs d_j[k] = pref[need + k] - pref[need], for k=0..(y[j]-need)
int maxk = min((ll)y[j] - need, (ll)v.size() - need);
extra[j].reserve(maxk + 1);
for(int k = 0; k <= maxk; k++){
extra[j].push_back(pref[need + k] - pref[need]);
}
// already sorted by construction
}
// merge function: K smallest sums of sorted A and B
auto mergeK = [&](const vector<ll> &A, const vector<ll> &B){
int n = A.size(), m = B.size();
vector<ll> C;
C.reserve(min((ll)K, (ll)n * m));
using T = tuple<ll,int,int>;
// min-heap via greater comparator
auto cmp = greater<T>();
priority_queue<T, vector<T>, decltype(cmp)> pq(cmp);
// push (A[i]+B[0], i, 0) for i=0..min(n-1, K-1)
int lim = min(n, K);
for(int i = 0; i < lim; i++){
pq.emplace(A[i] + B[0], i, 0);
}
while((int)C.size() < K && !pq.empty()){
auto [s, i, j] = pq.top();
pq.pop();
C.push_back(s);
if(j + 1 < m){
pq.emplace(A[i] + B[j+1], i, j+1);
}
}
return C;
};
// initial extra-sum list
vector<ll> L = {0};
// merge for each type
for(int j = 1; j <= M; j++){
if(extra[j].size() <= 1) continue; // only zero -> skip
L = mergeK(L, extra[j]);
}
// output K results
int produced = L.size();
for(int i = 0; i < produced; i++){
cout << (base_cost + L[i]) << '\n';
}
for(int i = produced; i < K; i++){
cout << -1 << '\n';
}
return 0;
}
