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>
#define mod 1000000007
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

struct edge {
	int to, cost;
};

int N, M, K, P[100000], C[100000], J[100000], p, q, r;
ll res[100000], dist[100000], dist2[100000];
vector<edge>v[100000];


int main() {
	cin >> N >> M >> K;
	for (int i = 0; i < N; ++i) {
		cin >> P[i];
	}
	for (int i = 0; i < N; ++i) {
		cin >> C[i];
		--C[i];
	}
	for (int i = 0; i < N; ++i) {
		cin >> J[i];
	}
	for (int i = 0; i < M; ++i) {
		cin >> p >> q >> r;
		--p;
		--q;
		v[p].push_back({ q,r });
		v[q].push_back({ p,r });
	}
	priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>>q;
	for (int i = 0; i < N; ++i) {
		dist[i] = -1;
		dist2[i] = -0xE869120E869120;
	}
	q.push(mkp(0, 0));
	while (!q.empty()) {
		int x = q.top().second;
		ll d = q.top().first;
		q.pop();
		if (dist[x] != -1)continue;
		dist[x] = d;
		for (edge e : v[x]) {
			if (dist[e.to] == -1) {
				q.push(mkp(d + e.cost, e.to));
			}
		}
	}
	for (int i = 0; i < N; ++i) {
		q.push(mkp(dist[i] - P[i], i));
	}
	while (!q.empty()) {
		int x = q.top().second;
		ll d = q.top().first;
		q.pop();
		if (dist2[x] != -0xE869120E869120)continue;
		dist2[x] = d;
		for (edge e : v[x]) {
			if (dist2[e.to] == -0xE869120E869120) {
				q.push(mkp(d + e.cost, e.to));
			}
		}
	}
	for (int i = 0; i < K; ++i) {
		res[i] = -0xE869120E869120;
	}
	for (int i = 0; i < N; ++i) {
		res[C[i]] = max(res[C[i]], J[i] - (dist[i] + dist2[i]));
	}
	for (int i = 0; i < K; ++i) {
		cout << res[i] << endl;
	}
}

