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 int ll;
typedef pair<ll,ll> mp;
ll inf =  1e12;


int main(){
    ll n,m,k;
    cin>>n>>m>>k;
    vector<ll> p(n),c(n),j(n);
    vector<vector<mp> > g(n);
    for(int i=0;i<n;i++)cin>>p[i];
    for(int i=0;i<n;i++)cin>>c[i];
    for(int i=0;i<n;i++)cin>>j[i];
    for(int i=0;i<m;i++){
	ll u,v,t;
	cin>>u>>v>>t;
	u--,v--;
	g[u].push_back( mp(t,v) );
	g[v].push_back( mp(t,u) );
    }
    vector<ll> dis(n,inf);
    dis[0] = 0;
    priority_queue<mp,vector<mp> ,greater<mp> > q;
    q.push( mp(0,0) );
    while( !q.empty() ){
	mp now = q.top();
	q.pop();
	ll nn = now.second;
	ll cost = now.first;
	if( dis[nn] != cost ) continue;
	for(int i=0;i<g[nn].size();i++){
	    ll next = g[nn][i].second;
	    ll ncost = cost + g[nn][i].first;
	    if( dis[next] <= ncost )continue;
	    dis[next] = ncost;
	    q.push( mp( ncost, next) );
	}
    }
    vector<ll> pan(n,-inf);
    //for(int i=0;i<n;i++)cout<<dis[i]<<endl;
    priority_queue<mp > que;
    for(int i=0;i<n;i++){
	pan[i] = p[i]-dis[i];
	que.push( mp(  p[i]  -dis[i], i) );
    }
    while( !que.empty() ){
	mp now = que.top();
	que.pop();
	ll nn = now.second;
	ll cost = now.first;
	if( pan[nn] > cost ) continue;
	for(int i=0;i<g[nn].size();i++){
	    ll next = g[nn][i].second ;
	    ll ncost = cost - g[nn][i].first ;
	    if( pan[next] >= ncost )continue;
	    pan[next] = ncost;
	    que.push( mp( ncost, next) );
	}
    }
    vector<ll> ans(k,-inf);
    for(int i=0;i<n;i++){
	//cout<<pan[i]<<' '<<dis[i]<<' '<<i<<endl;
	ans[c[i]-1] = max( ans[c[i]-1] , pan[i] - dis[i] + j[i] ) ;
    }

    for(int i=0;i<k;i++)cout<<ans[i]<<endl;
    

    return 0;
}

