#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = (1LL << 60);
// ---- Persistent Segment Tree for point-add & range-sum ----
struct PST {
struct Node { int l, r, sum; };
vector<Node> st;
int N;
PST(int _N): N(_N) {
st.reserve(_N * 20);
st.push_back({0,0,0}); // dummy node at idx=0
}
// build empty tree on [l..r], returns root index
int build(int l, int r) {
int id = st.size();
st.push_back({0,0,0});
if (l < r) {
int m = (l + r) >> 1;
st[id].l = build(l, m);
st[id].r = build(m+1, r);
}
return id;
}
// create new version from prev, add v at pos p
int update(int prev, int l, int r, int p, int v) {
int id = st.size();
st.push_back(st[prev]);
if (l == r) {
st[id].sum += v;
} else {
int m = (l + r) >> 1;
if (p <= m)
st[id].l = update(st[prev].l, l, m, p, v);
else
st[id].r = update(st[prev].r, m+1, r, p, v);
st[id].sum = st[st[id].l].sum + st[st[id].r].sum;
}
return id;
}
// query sum on [ql..qr]
int query(int id, int l, int r, int ql, int qr) const {
if (!id || qr < l || r < ql) return 0;
if (ql <= l && r <= qr) return st[id].sum;
int m = (l + r) >> 1;
return query(st[id].l, l, m, ql, qr)
+ query(st[id].r, m+1, r, ql, qr);
}
};
// solve function to be called by the judge
// signature matches: solve(int n, int m, int w,
// vector<int> x, vector<int> y,
// vector<int> a, vector<int> b, vector<int> c,
// vector<int> l, vector<int> r, vector<int> t)
long long solve(int n, int m, int w,
vector<int> x, vector<int> y,
vector<int> a, vector<int> b, vector<int> c,
vector<int> L, vector<int> R, vector<int> T) {
// 1) collect all time points for coordinate compression
vector<int> allT;
allT.reserve(2*m + 2*w + 1);
allT.push_back(0);
for(int i = 0; i < m; i++){
allT.push_back(a[i]);
allT.push_back(b[i]);
}
for(int i = 0; i < w; i++){
allT.push_back(L[i]);
allT.push_back(R[i]);
}
sort(allT.begin(), allT.end());
allT.erase(unique(allT.begin(), allT.end()), allT.end());
auto comp = [&](int v) {
return int(lower_bound(allT.begin(), allT.end(), v) - allT.begin());
};
int Tsz = allT.size();
// 2) bucket meals by compressed R
vector<vector<int>> meals_at_r(Tsz);
for(int i = 0; i < w; i++){
int lc = comp(L[i]);
int rc = comp(R[i]);
meals_at_r[rc].push_back(lc);
}
// 3) build persistent segment tree versions
PST pst(Tsz);
vector<int> version(Tsz);
version[0] = pst.build(0, Tsz-1);
for(int lc: meals_at_r[0]){
version[0] = pst.update(version[0], 0, Tsz-1, lc, 1);
}
for(int i = 1; i < Tsz; i++){
version[i] = version[i-1];
for(int lc: meals_at_r[i]){
version[i] = pst.update(version[i], 0, Tsz-1, lc, 1);
}
}
// helper to get g(b, a): #meals completely inside (b, a)
auto get_g = [&](int b, int a) {
int bc = comp(b), ac = comp(a);
if (ac == 0) return 0;
int total = pst.query(version[ac-1], 0, Tsz-1, 0, Tsz-1);
int bad = pst.query(version[ac-1], 0, Tsz-1, 0, bc);
return total - bad;
};
// 4) prepare train indices sorted by departure time
vector<int> idx(m);
iota(idx.begin(), idx.end(), 0);
sort(idx.begin(), idx.end(),
[&](int i, int j){ return a[i] < a[j]; });
// dp[i]: min cost to finish train i
vector<ll> dp(m, INF);
// for each planet, a min-heap of (arrival_time b_j, train_index j)
vector< priority_queue<pair<int,int>,
vector<pair<int,int>>, greater<pair<int,int>>> > heap(n);
// and a deque to maintain candidate js with decision-monotonicity
vector< deque<int> > dq(n);
// initial state: planet 0 at time 0, treat j=-1 with dp=-0
heap[0].push({0, -1});
auto get_dp = [&](int j){
return j < 0 ? 0LL : dp[j];
};
auto get_b = [&](int j){
return j < 0 ? 0 : b[j];
};
ll answer = INF;
// 5) main scan in order of departure time
for(int _k = 0; _k < m; _k++){
int i = idx[_k];
int p = x[i];
int ai = a[i];
// step1: move all arriving-by-ai trains from heap[p] into dq[p]
auto &hp = heap[p];
auto &dq_p = dq[p];
while(!hp.empty() && hp.top().first <= ai){
int j = hp.top().second;
hp.pop();
// pop-back by quadrangle-inequality check
while(dq_p.size() >= 2){
int j2 = dq_p.back();
int j1 = dq_p[dq_p.size()-2];
ll lhs = get_dp(j2)
+ (ll)get_g(get_b(j2), ai) * T[p];
ll rhs = get_dp(j1)
+ (ll)get_g(get_b(j1), ai) * T[p];
if (lhs >= rhs) dq_p.pop_back();
else break;
}
dq_p.push_back(j);
}
// step2: front of dq[p] is best predecessor
if (!dq_p.empty()){
int j = dq_p.front();
dp[i] = get_dp(j)
+ (ll)get_g(get_b(j), ai) * T[p]
+ c[i];
if (y[i] == n-1)
answer = min(answer, dp[i]);
}
// step3: push this train into heap at its arrival planet
heap[y[i]].push({ b[i], i });
}
if (answer >= INF/2) return -1;
return answer;
}
