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 <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <complex>
#include <array>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
typedef vector<double> VD;
typedef vector<VD> VVD;

template<typename T> void chmin(T &a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T &a, T b) { if (a < b) a = b; }

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }

const ll INF = 1e17;

struct edge{ll to,cost;};
 
void dijkstra(vector<vector<edge> > &graph, int start, VL &d){
    int n = graph.size();
    d.assign(n, INF);
    d[start] = 0;
    priority_queue<PL, vector<PL>, greater<PL> > q;
    q.push(make_pair(0, start));
    while (!q.empty()){
        PL a = q.top();
        q.pop();
        int now = a.second;
        if(d[now] < a.first) continue;
        REP(i,graph[now].size()){
            edge e = graph[now][i];
            int next = e.to;
            if (d[next] > d[now] + e.cost){
                d[next] = d[now] + e.cost;
                q.push(make_pair(d[next], next));
            }
        }
    }
}

int main() {
    int n, m, k;
    cin >> n >> m >> k;
    VI p(n), c(n), j(n);
    REP(i,n) p[i] = in();
    REP(i,n) c[i] = in() - 1;
    REP(i,n) j[i] = in();

    vector<vector<edge>> e(n), e2(2 * n);
    REP(i,m){
        int u = in() - 1, v = in() - 1, t = in();
        e[u].push_back(edge{v, t});
        e[v].push_back(edge{u, t});
        // e2[u].push_back(edge{v, t});
        // e2[v].push_back(edge{u, t});
        e2[n + u].push_back(edge{n + v, t});
        e2[n + v].push_back(edge{n + u, t});
    }
    // REP(i,n) e2[i].push_back(edge{n + i, -p[i]});
    VL dist, cost;
    dijkstra(e, 0, dist);
    REP(i,n) e2[0].push_back(edge{n + i, dist[i] - p[i]});
    dijkstra(e2, 0, cost);

    // REP(i,n) cout << dist[i] << " ";
    // cout << endl;
    // REP(i,2*n) cout << cost[i] << " ";
    // cout << endl;

    VL ans(k, -INF);
    REP(i,n){
        ll tmp = j[i] - dist[i] - cost[n + i];
        chmax(ans[c[i]], tmp);
    }
    REP(i,k) printf("%lld\n", ans[i]);

    return 0;
}

