E: Jam

problem

There are N cities in a country, numbered 1, \ 2, \ ..., \ N. These cities are connected in both directions by M roads, and the i-th road allows you to travel between the cities u_i and v_i in time t_i. Also, any two cities can be reached by using several roads.

Bread is sold in each town, and the deliciousness of bread sold in town i is P_i.

In addition, there are K flavors of jam in this country, and in town i you can buy delicious J_i jams at taste c_i.

Homu-chan, who lives in Town 1, decided to go out to buy bread and jam one by one. Homu-chan travels through several cities, buys bread and jam, and returns to city 1. More precisely, go to city u to buy bread, select city v to buy jam, and return to city 1. At this time, u = v, u = 1, and v = 1 can be used.

The happiness of Homu-chan after shopping is "the total of the deliciousness of the bread and jam I bought"-"the time it took to move". For each of the K types of jam, calculate the maximum possible happiness when going to buy jam and bread of that taste.

Input format


NMK
P_1 P_2 ... P_N
c_1 c_2 ... c_N
J_1 J_2 ... J_N
u_1 v_1 t_1
u_2 v_2 t_2
::
u_M v_M t_M


Constraint

* 1 \ leq N \ leq 10 ^ 5
* 1 \ leq M \ leq 2 \ times 10 ^ 5
* 1 \ leq K \ leq N
* 1 \ leq P_i, J_i \ leq 10 ^ 9
* 1 \ leq c_i \ leq K
* 1 \ leq u_i, v_i \ leq N
* 1 \ leq t_i \ leq 10 ^ 9
* For all i \ (1 \ leq i \ leq K), there exists j \ (1 \ leq j \ leq N) such that c_j = i.
* Given graphs are concatenated graphs that do not include multiple edges or self-loops.
* All inputs are given as integers



Output format

Please output K line. On line i, print the maximum happiness when you go to buy bread and taste i jam.

Input example 1


4 4 3
3 6 1 6
1 1 2 3
6 1 5 5
1 2 1
2 3 1
1 3 1
1 4 1


Output example 1


Ten
8
9


* When buying taste 1 jam
* If you buy jam in city 1, move to city 2, buy bread, and come back to city 1, your happiness will be 6 + 6-2 = 10, which is optimal.
* When buying taste 2 jam
* If you move to City 2 to buy bread, move to City 3 to buy jam, and return to City 1, your happiness will be 6 + 5-3 = 8.
* When buying taste 3 jam
* If you go to City 4 and buy both bread and jam and come back to City 1, your happiness will be 6 + 5-2 = 9.



Input example 2


2 1 2
1 1
1 2
1 1
1 2 1000000000


Output example 2


2
-1999999998


* Homu-chan seems to be dissatisfied because the distance is too far compared to the deliciousness of bread and jam.



Input example 3


6 8 3
31 41 59 26 53 58
1 2 3 1 2 3
27 18 28 18 28 46
one two Three
2 3 7
2 4 8
3 6 9
3 5 3
4 5 3
1 5 15
4 6 7


Output example 3


66
61
68






Example

Input

4 4 3
3 6 1 6
1 1 2 3
6 1 5 5
1 2 1
2 3 1
1 3 1
1 4 1


Output

10
8
9
#include <bits/stdc++.h>
using namespace std;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

struct Edge { int to, cost; };

template<class T, class Edge> V<T> dijkstra(const VV<Edge>& g, int s = 0) {
  V<T> dist(g.size(), numeric_limits<T>::max());
  using P = pair<T, int>;
  priority_queue< P, V<P>, greater<P> > pque;
  pque.emplace(dist[s] = 0, s);
  while (!pque.empty()) {
    T d; int v;
    tie(d, v) = pque.top(); pque.pop();
    if (d > dist[v]) continue;
    for (const auto& e : g[v]) if (dist[v] + e.cost < dist[e.to]) {
      pque.emplace(dist[e.to] = dist[v] + e.cost, e.to);
    } 
  }
  return dist;
}

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n, m, k; cin >> n >> m >> k;
  V<> p(n); for (auto&& e : p) cin >> e;
  V<> c(n); for (auto&& e : c) cin >> e, --e;
  V<> q(n); for (auto&& e : q) cin >> e;
  VV<Edge> g(2 * n);
  while (m--) {
    int u, v, c; cin >> u >> v >> c, --u, --v;
    g[u].push_back(Edge{v, c});
    g[v].push_back(Edge{u, c});
    g[n + u].push_back(Edge{n + v, c});
    g[n + v].push_back(Edge{n + u, c});
  }
  constexpr int o = 1e9;
  for (int v = 0; v < n; ++v) {
    g[v].push_back(Edge{n + v, o - p[v]});
  }
  auto d = dijkstra<lint>(g);
  V<lint> res(k, -9e18);
  for (int v = 0; v < n; ++v) {
    res[c[v]] = max(res[c[v]], o + q[v] - d[n + v] - d[v]);
  }
  for (lint e : res) cout << e << '\n';
}
