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;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

struct FastIO{
  FastIO(){
    cin.tie(0);
    ios::sync_with_stdio(0);
  }
}fastio_beet;


template <typename T>
vector<T> dijkstra(Int s,vector<vector<pair<Int, T> > > & G){
  const T INF = numeric_limits<T>::max();
  using P = pair<T, Int>;
  Int n=G.size();
  vector<T> d(n,INF);
  vector<Int> b(n,-1);
  priority_queue<P,vector<P>,greater<P> > q;
  d[s]=0;
  q.emplace(d[s],s);
  while(!q.empty()){
    P p=q.top();q.pop();
    Int v=p.second;
    if(d[v]<p.first) continue;
    for(auto& e:G[v]){
      Int u=e.first;
      T c=e.second;
      if(d[u]>d[v]+c){
        d[u]=d[v]+c;
        b[u]=v;
        q.emplace(d[u],u);
      }
    }
  }
  return d;
}

//INSERT ABOVE HERE
signed main(){
  Int n,m,k;
  cin>>n>>m>>k;
  vector<Int> ps(n),cs(n),js(n);
  for(Int i=0;i<n;i++) cin>>ps[i];
  for(Int i=0;i<n;i++) cin>>cs[i],cs[i]--;
  for(Int i=0;i<n;i++) cin>>js[i];

  using P = pair<Int, Int>;
  vector< vector<P> > G(n+1);
  for(Int i=0;i<m;i++){
    Int u,v,t;
    cin>>u>>v>>t;
    u--;v--;
    G[u].emplace_back(v,t);
    G[v].emplace_back(u,t);
  }

  auto ds=dijkstra(0,G);

  const Int OFS = 2e9;
  for(Int i=0;i<n;i++)
    G[n].emplace_back(i,ds[i]-ps[i]+OFS);

  auto dt=dijkstra(n,G);
  const Int INF = 1e18;
  vector<Int> ans(k,-INF);
  for(Int i=0;i<n;i++)
    chmax(ans[cs[i]],js[i]-(dt[i]-OFS+ds[i]));

  for(Int x:ans) cout<<x<<"\n";
  cout<<flush;
  return 0;
}

