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, int> P;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);

#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	a = (a<MOD)?a:(a-MOD);
	b = (b<MOD)?b:(b-MOD);
	a += b;
	a = (a<MOD)?a:(a-MOD);
}

ll ans[100100],jm[100100],dist[2][225816];
int c[100100];
vector<P> g[225816];
priority_queue<P, vector<P>, greater<P> > pq;

int main(void){
	int n,m,k,i,j,u,v;
	ll p,t;
	fill(ans,ans+100100,-INFLL);
	fill(dist[0],dist[2],INFLL);
	cin >> n >> m >> k;
	for(i=0; i<n; ++i){
		cin >> p;
		g[i+n].push_back(P(MOD-p,i));
	}
	for(i=0; i<n; ++i){
		cin >> c[i]; --c[i];
	}
	for(i=0; i<n; ++i){
		cin >> jm[i];
	}
	for(i=0; i<m; ++i){
		cin >> u >> v >> t;
		--u; --v;
		g[u].push_back(P(t,v));
		g[v].push_back(P(t,u));
		g[u+n].push_back(P(t,v+n));
		g[v+n].push_back(P(t,u+n));
	}
	for(j=0; j<2; ++j){
		dist[j][n*j] = 0ll;
		pq.push(P(dist[j][n*j],n*j));
		while(!pq.empty()){
			u = pq.top().second;
			t = pq.top().first;
			pq.pop();
			if(t > dist[j][u]){
				continue;
			}
			for(i=(g[u].l_ength()-1); i>=0; --i){
				v = g[u][i].second;
				t = g[u][i].first;
				if(dist[j][v] > dist[j][u] + t){
					dist[j][v] = dist[j][u] + t;
					pq.push(P(dist[j][v],v));
				}
			}
		}
	}
	for(i=0; i<n; ++i){
		ans[c[i]] = max(ans[c[i]],jm[i]+MOD-dist[0][i]-dist[1][i]);
	}
	for(i=0; i<k; ++i){
		cout << ans[i] << endl;
	}
	return 0;
}


