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 rep(i,n) for (int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<pi, pi> pp;
typedef pair<ll, ll> pl;
const double EPS = 1e-9;
const ll MOD = 1000000007;
const int inf = 1 << 30;
const ll linf = 1LL << 60;

int n, m, k;
ll P[100000], C[100000], J[100000];
vector<pl> e[100000];
ll dist[2][100000];
ll ans[100000];
priority_queue<pl, vector<pl>, greater<pl>> que, que2;

int main() {
    cin >> n >> m >> k;
    rep(i,k) ans[i] = -linf;
    rep(i,n) cin >> P[i];
    rep(i,n) cin >> C[i];
    rep(i,n) C[i]--;
    rep(i,n) cin >> J[i];
    rep(i,m) {
        int u, v, t;
        cin >> u >> v >> t; u--; v--;
        e[u].push_back(pl(v, t));
        e[v].push_back(pl(u,t));
    }
    rep(i,2) rep(j,n) dist[i][j] = linf;
    dist[0][0] = 0; dist[1][0] = -P[0];
    que.push(pl(0, 0));
    que2.push(pl(-P[0], n));
    while (!que.empty()) {
        pl q = que.top(); que.pop();
        ll f = q.second/n, pos = q.second%n;
        if (dist[f][pos] < q.first) continue;
        rep(i,e[pos].size()) {
            pl to = e[pos][i];
            if (f == 1) {
                /*
                if (dist[1][to.first] > to.second + dist[f][pos]) {
                    dist[1][to.first] = to.second + dist[f][pos];
                    que.push(pl(dist[1][to.first], n+to.first));
                }
                */
            } else {
                if (dist[0][to.first] > to.second + dist[f][pos]) {
                    dist[0][to.first] = to.second + dist[f][pos];
                    que.push(pl(dist[0][to.first], to.first));
                }
                if (dist[1][to.first] > to.second + dist[f][pos] - P[to.first]) {
                    dist[1][to.first] = to.second + dist[f][pos] - P[to.first];
                    que2.push(pl(dist[1][to.first], n+to.first));
                }
            }
        }
    }
    while (!que2.empty()) {
        pl q = que2.top(); que2.pop();
        ll f = q.second/n, pos = q.second%n;
        if (dist[f][pos] < q.first) continue;
        rep(i,e[pos].size()) {
            pl to = e[pos][i];
            if (f == 1) {
                if (dist[1][to.first] > to.second + dist[f][pos]) {
                    dist[1][to.first] = to.second + dist[f][pos];
                    que2.push(pl(dist[1][to.first], n+to.first));
                }
            }
        }
    }
    rep(i,n) {
        //cerr << dist[0][i] << " " << dist[1][i] << " " << dist[2][i] << endl;
        //ans[C[i]] = max(ans[C[i]], -1 * (dist[1][i] +dist[2][i]));
        ans[C[i]] = max(ans[C[i]], -1 * (dist[1][i] +dist[0][i] -J[i]));
    }
    rep(i,k) {
        cout << ans[i] << endl;
    }
}

