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 <queue>
#include <string>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const long long inf = 1LL << 60;
const long long limit = 1500000000;
struct edge {
    int to; long long cost;
};
struct state {
    int pos, tp; long long cost;
};
bool operator<(const state& s1, const state& s2) {
    return s1.cost != s2.cost ? s1.cost > s2.cost : s1.pos * 2 + s1.tp < s2.pos * 2 + s2.tp;
}
long long dist[100003][2];
bool visited[100003][2];
vector<edge> G[100003];
int main() {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    int N, M, K;
    cin >> N >> M >> K;
    vector<int> P(N), C(N), J(N);
    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) {
        int a, b, c;
        cin >> a >> b >> c; --a, --b;
        G[a].push_back(edge{ b, c });
        G[b].push_back(edge{ a, c });
    }
    for (int i = 0; i < N; ++i) {
        dist[i][0] = dist[i][1] = inf;
    }
    dist[0][0] = 0;
    priority_queue<state> que;
    que.push(state{ 0, 0, 0 });
    while (!que.empty()) {
        state u = que.top(); que.pop();
        if(visited[u.pos][u.tp]) continue;
        visited[u.pos][u.tp] = true;
        long long d = dist[u.pos][u.tp];
        for (edge e : G[u.pos]) {
            if (dist[e.to][u.tp] > d + e.cost) {
                dist[e.to][u.tp] = d + e.cost;
                que.push(state{ e.to, u.tp, dist[e.to][u.tp] });
            }
        }
        if (u.tp == 0 && dist[u.pos][1] > d + limit - P[u.pos]) {
            dist[u.pos][1] = d + limit - P[u.pos];
            que.push(state{ u.pos, 1, dist[u.pos][1] });
        }
    }
    vector<long long> ans(K, -inf);
    for (int i = 0; i < N; ++i) {
        ans[C[i]] = max(ans[C[i]], limit - (dist[i][0] + dist[i][1]) + J[i]);
    }
    for (int i = 0; i < K; ++i) {
        cout << ans[i] << '\n';
    }
    return 0;
}

