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 vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define REP(i,n) for(int i=0;i<(int)(n);i++)
#define FOR(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define FORR(i,a,b) for(int i=(int)(b)-1;i>=(int)(a);i--)

#define CHMIN(a,b) (a)=min((a),(b))
#define CHMAX(a,b) (a)=max((a),(b))

#define DEBUG(x) cout<<#x<<": "<<(x)<<endl

int n,m,k;
int p[125252];
int c[125252];
int j[125252];

vl dijkstra(vector<vector<pll>> &g, int s){
	int n = g.size();
	vl dist(n,1ll<<60);
	priority_queue<pll> Q;
	dist[s] = 0;
	Q.push(pll(-0,s));
	while(Q.size()){
		pll P = Q.top(); Q.pop();
		int p = P.second;
		if(dist[p] != -P.first)continue;
		for(pll PP : g[p]){
			int to = PP.first;
			ll cost = PP.second;
			ll nd = dist[p] + cost;
			if(nd < dist[to]){
				dist[to] = nd;
				Q.push(pll(-nd,to));
			}
		}
	}
	return dist;
}

int main(){
	scanf("%d%d%d",&n,&m,&k);
	REP(i,n)scanf("%d",p+i);
	REP(i,n)scanf("%d",c+i),c[i]--;
	REP(i,n)scanf("%d",j+i);
	// 0 - i
	vector<vector<pll>> g(n);
	REP(i,m){
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		--x;--y;
		g[x].push_back(pll(y,z));
		g[y].push_back(pll(x,z));
	}
	vl dist = dijkstra(g,0);
	// 0 - pan - i
	vector<vector<pll>> g2(n+n);
	REP(i,n)for(pll P : g[i]){
		g2[i].push_back(P);
	}
	REP(i,n){
		g2[i].push_back(pll(i+n,-p[i]));
	}
	vl dist2 = dijkstra(g2,0);
	// (0-pan-) i - jam (-0)
	vector<vector<pll>> g3(1+n+k);
	REP(i,n)for(pll P : g[i]){
		g3[i].push_back(P);
	}
	REP(i,n){
		g3[n+k].push_back(pll(i,dist2[i+n]));
		g3[i].push_back(pll(n+c[i],-j[i]+dist[i]));
	}
	vl dist3 = dijkstra(g3,n+k);
	// output
	REP(i,k){
		printf("%lld\n",-dist3[n+i]);
	}
	return 0;
}
