#include <bits/stdc++.h>
using namespace std;
struct Coupon {
int idx; // original index
long long p; // price
int t; // type (multiplier)
};
/* comparison for coupons with T > 1:
increasing K = p * t / (t-1)
(compare a.p * a.t * (b.t-1) with b.p * b.t * (a.t-1) )
*/
bool cmpK(const Coupon& a, const Coupon& b) {
// both a.t > 1 and b.t > 1
__int128 left = (__int128)a.p * a.t * (b.t - 1);
__int128 right = (__int128)b.p * b.t * (a.t - 1);
if (left != right) return left < right;
// tie‑break – any deterministic rule is fine
if (a.t != b.t) return a.t > b.t; // larger multiplier first
return a.p > b.p; // larger price first
}
/* ------------------------------------------------------------ */
vector<int> max_coupons(int A, vector<int> P, vector<int> T) {
const int N = (int)P.size();
/* split coupons */
vector<Coupon> pos, ones;
for (int i = 0; i < N; ++i) {
if (T[i] > 1)
pos.push_back({i, P[i], T[i]});
else
ones.push_back({i, P[i], T[i]}); // T == 1
}
/* sort as proved in the analysis */
sort(pos.begin(), pos.end(), cmpK); // increasing K
sort(ones.begin(), ones.end(),
[](const Coupon& a, const Coupon& b) {
return a.p < b.p; // cheapest first
});
vector<Coupon> all;
all.reserve(N);
all.insert(all.end(), pos.begin(), pos.end());
all.insert(all.end(), ones.begin(), ones.end());
const int M = (int)all.size(); // M == N
/* DP */
const __int128 INF = (__int128)1 << 120; // huge enough
vector<vector<__int128>> dp(M + 1, vector<__int128>(M + 1, -1));
vector<vector<char>> take(M + 1, vector<char>(M + 1, 0));
dp[0][0] = (__int128)A;
for (int i = 1; i <= M; ++i) {
const Coupon& c = all[i - 1];
dp[i][0] = dp[i - 1][0];
take[i][0] = 0;
for (int k = 1; k <= i; ++k) {
/* skip */
dp[i][k] = dp[i - 1][k];
take[i][k] = 0;
/* take */
if (dp[i - 1][k - 1] != -1 && dp[i - 1][k - 1] >= c.p) {
__int128 base = dp[i - 1][k - 1] - c.p;
__int128 after;
if (base > INF / c.t) after = INF;
else after = base * (__int128)c.t;
if (after > dp[i][k]) {
dp[i][k] = after;
take[i][k] = 1;
}
}
}
}
/* best obtainable number of coupons */
int best = 0;
for (int k = M; k >= 0; --k) {
if (dp[M][k] != -1) {
best = k;
break;
}
}
/* reconstruction */
vector<int> answer;
int i = M, k = best;
while (i > 0) {
if (take[i][k]) {
answer.push_back(all[i - 1].idx);
--k;
}
--i;
}
reverse(answer.begin(), answer.end());
return answer;
}
