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;

typedef pair<int64_t, int64_t> P;
const int64_t INF = 1e18;

void dijkstra(int N, vector<P> edges[], vector<int64_t>& dist){
    priority_queue<P, vector<P>, greater<P>> que;
    for(int i=0; i<N; i++) que.push({dist[i], i});
    while(que.size()){
        auto p = que.top(); que.pop();
        int i = p.second;
        int64_t d = p.first;
        if(d > dist[i]) continue;
        for(auto& e : edges[i]){
            int j = e.first;
            int64_t d2 = d + e.second;
            if(d2 < dist[j]){
                dist[j] = d2;
                que.push({d2, j});
            }
        }
    }
}

vector<P> edges[100000];

int main(){
    int N, M, K;
    cin >> N >> M >> K;
    int P[100000], C[100000], J[100000];
    for(int i=0; i<N; i++) cin >> P[i];
    for(int i=0; i<N; i++) cin >> 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;
        edges[a-1].emplace_back(b-1, c);
        edges[b-1].emplace_back(a-1, c);
    }

    vector<int64_t> D0(N, INF);
    D0[0] = 0;
    dijkstra(N, edges, D0);

    vector<int64_t> cost(N);
    for(int i=0; i<N; i++) cost[i] = D0[i] - P[i];
    dijkstra(N, edges, cost);

    vector<int64_t> ans(K+1, -INF);
    for(int i=0; i<N; i++){
        int64_t res = J[i] - cost[i] - D0[i];
        ans[C[i]] = max(ans[C[i]], res);
    }
    for(int i=1; i<=K; i++) cout << ans[i] << endl;
    return 0;
}

