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 <iostream>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <tuple>
#include <bitset>
#include <memory>
#include <cmath>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <numeric>
#include <climits>
#include <cfloat>

struct Edge {
	int to, cost;
};
int main() {
	int n, m, k; std::cin >> n >> m >> k;
	std::vector<int> bread_taste(n), jam_taste(n);
	std::vector<std::vector<int>> jam_type(k);
	for (auto& p : bread_taste) std::cin >> p;
	for (auto i = 0; i < n; ++i) {
		int c; std::cin >> c;
		jam_type[c - 1].push_back(i);
	}
	for (auto& j : jam_taste) std::cin >> j;
	std::vector<std::vector<Edge>> towns(n);
	for (auto i = 0; i < m; ++i) {
		int u, v, t; std::cin >> u >> v >> t;
		towns[u - 1].push_back({ v - 1, t });
		towns[v - 1].push_back({ u - 1, t });
	}
	auto comparator = [](const std::pair<int, long long int>& a, const std::pair<int, long long int>& b) {return a.second > b.second; };
	std::priority_queue<std::pair<int, long long int>, std::vector<std::pair<int, long long int>>, decltype(comparator)> queue(comparator);
	std::vector<long long int> min_time_from_start(n, LLONG_MAX); min_time_from_start[0] = 0; queue.emplace(0, 0);
	while (!queue.empty()) {
		auto top = queue.top(); queue.pop();
		if (min_time_from_start[top.first] == top.second) {
			for (const auto& e : towns[top.first]) {
				if (min_time_from_start[e.to] > top.second + e.cost) {
					min_time_from_start[e.to] = top.second + e.cost;
					queue.emplace(e.to, top.second + e.cost);
				}
			}
		}
	}
	std::vector<long long int> min_cost_for_bread(n);
	for (auto i = 0; i < n; ++i) min_cost_for_bread[i] = min_time_from_start[i] - bread_taste[i];
	std::vector<long long int> min_cost_for_jam(min_cost_for_bread);
	for (auto i = 0; i < n; ++i) queue.emplace(i, min_cost_for_jam[i]);
	while (!queue.empty()) {
		auto top = queue.top(); queue.pop();
		if (min_cost_for_jam[top.first] == top.second) {
			for (const auto& e : towns[top.first]) {
				if (min_cost_for_jam[e.to] > top.second + e.cost) {
					min_cost_for_jam[e.to] = top.second + e.cost;
					queue.emplace(e.to, top.second + e.cost);
				}
			}
		}
	}
	for (auto i = 0; i < n; ++i) min_cost_for_jam[i] -= jam_taste[i];
	std::vector<long long int> min_cost_for_all(min_time_from_start);
	for (auto i = 0; i < n; ++i) min_cost_for_all[i] += min_cost_for_jam[i];
	for (auto type = 0; type < k; ++type) {
		std::cout << -min_cost_for_all[*std::min_element(jam_type[type].begin(), jam_type[type].end(), [&min_cost_for_all](const int i, const int j) {return min_cost_for_all[i] < min_cost_for_all[j]; })] << std::endl;
	}
}
