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 <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, M, K;
	cin >> N >> M >> K;
	vector<int> P(N);
	vector<int> C(N);
	vector<int> J(N);
	int mx = 0;
	for (int i = 0; i < N; i++) {
		cin >> P[i];
		mx = max(mx, P[i]);
	}
	mx++;
	for (int i = 0; i < N; i++) {
		cin >> C[i];
	}
	for (int i = 0; i < N; i++) {
		cin >> J[i];
	}
	vector<vector<int> > g(N);
	vector<vector<int> > d(N);
	int a, b, c;
	for (int i = 0; i < M; i++) {
		cin >> a >> b >> c; a--; b--;
		g[a].push_back(b);
		g[b].push_back(a);
		d[a].push_back(c);
		d[b].push_back(c);
	}
	int INF = (int)1 << 60;
	vector<vector<int> > dist(N, vector<int>(2, INF));
	dist[0][0] = 0;
	dist[0][1] = mx - P[0];

	priority_queue<pair<int, pair<int, int> > > pq;
	pq.emplace(-dist[0][0], make_pair(0, 0));
	pq.emplace(-dist[0][1], make_pair(0, 1));

	while (pq.size()) {
		int v = pq.top().second.first;
		int t = pq.top().second.second;
		if (dist[v][t] != -pq.top().first) {
			pq.pop();
			continue;
		}
		pq.pop();
		for (int i = 0; i < g[v].size(); i++) {
			if (dist[g[v][i]][t] > dist[v][t] + d[v][i]) {
				dist[g[v][i]][t] = dist[v][t] + d[v][i];
				pq.emplace(-dist[g[v][i]][t], make_pair(g[v][i], t));
			}
			if (t == 0) {
				int val = dist[v][0] + d[v][i] + mx - P[g[v][i]];
				if (dist[g[v][i]][1] > val) {
					dist[g[v][i]][1] = val;
					pq.emplace(-dist[g[v][i]][1], make_pair(g[v][i], 1));
				}
			}
		}
	}
	vector<int> res(K, -INF);

	for (int i = 0; i < N; i++) {
		res[C[i] - 1] = max(res[C[i] - 1], J[i] + mx - dist[i][1] - dist[i][0]);
	}
	for (int i = 0; i < K; i++) {
		cout << res[i] << endl;
	}
}
