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 long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;
template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

#define EPS (1e-7)
const ll INF = 1e18;
#define PI (acos(-1))
//const ll mod = 1000000007;
//1からDijkstra パンを買うまでの距離を記録
//パンの値段も考慮した初期距離で、多点始点Dijkstra

ll N, M, K;
ll P[105000];
ll c[105000], J[105000];
vector<l_l> pathes[105000];
priority_queue<l_l, vector<l_l>, greater<l_l>> que;

ll dist[2][105000];

void Dijkstra(ll index) {
    while(!que.empty()) {
        l_l now = que.top();
        que.pop();
        if(dist[index][now.second] < now.first) continue;
        for(int i = 0; i < pathes[now.second].size(); i++) {
            ll to = pathes[now.second][i].first;
            ll newcost = now.first + pathes[now.second][i].second;
            if(chmin(dist[index][to], newcost)) {
                que.push({newcost, to});
            }
        }
    }
}

ll ans[105000];

int main() {
    //cout.precision(10);
    cin >> N >> M >> K;
    for(int i = 1; i <= N; i++) cin >> P[i];
    for(int i = 1; i <= N; i++) cin >> c[i];
    for(int i = 1; i <= N; i++) cin >> J[i];
    for(int i = 1; i <= M; i++) {
        ll u, v, t;
        cin >> u >> v >> t;
        pathes[u].push_back({v, t});
        pathes[v].push_back({u, t});
    }
    for(int i = 1; i <= N; i++) {
        dist[0][i] = INF;
    }
    dist[0][1] = 0;
    que.push({0, 1});
    Dijkstra(0);
    for(int i = 1; i <= N; i++) {
        dist[1][i] = dist[0][i] - P[i];
        que.push({dist[1][i], i});
    }
    Dijkstra(1);
    for(int i = 1; i <= K; i++) ans[i] = INF;
    for(int i = 1; i <= N; i++) {
        chmin(ans[c[i]], dist[0][i] - J[i] + dist[1][i]);
    }
    for(int i = 1; i <= K; i++) cout << -ans[i] << endl;
    return 0;
}

