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
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// MMMMM#<<_jwr:(Z4QHHMMHMHHHHHHHHHHHHHHHHHHHHHHHHHHHH###NHSXZ>` ~````.OXtt>~._<?MM
*/

#include <bits/stdc++.h>
using namespace std;

using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;
using LL = long long;
using VL = vector<LL>;
using VVL = vector<VL>;
using PLL = pair<LL, LL>;
using VS = vector<string>;

#define ALL(a)  begin((a)),end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define MT make_tuple
#define SZ(a) int((a).size())
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))

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

#define FF first
#define SS second

template<class S, class T>
istream& operator>>(istream& is, pair<S,T>& p){
  return is >> p.FF >> p.SS;
}
template<class T>
istream& operator>>(istream& is, vector<T>& xs){
  for(auto& x: xs)
	is >> x;
  return is;
}
template<class S, class T>
ostream& operator<<(ostream& os, const pair<S,T>& p){
  return os << p.FF << " " << p.SS;
}
template<class T>
ostream& operator<<(ostream& os, const vector<T>& xs){
  for(unsigned int i=0;i<xs.size();++i)
	os << (i?" ":"") << xs[i];
  return os;
}
template<class T>
void maxi(T& x, T y){
  if(x < y) x = y;
}
template<class T>
void mini(T& x, T y){
  if(x > y) x = y;
}
void debug(istringstream&){}
template <char sep=',', class Head, class... Tail>
void debug(istringstream& iss, Head&& head, Tail&&... tail)
{
  string name;
  getline(iss, name, ',');
  cout << sep << name << "=" << head;
  debug(iss, forward<Tail>(tail)...);
}

#ifdef PYONPOI
#define DEBUG(...) \
  do{ \
    istringstream ss(#__VA_ARGS__); \
    debug<' '>(ss, __VA_ARGS__); \
    cout<<endl; \
  }while(0)
#else
#define DEBUG
#endif


const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL MOD = 1e9+7;
const LL INF = 1e18;

int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  int N, M, K;
  cin >> N >> M >> K;
  VL pan(N), kind(N), jam(N);
  cin >> pan >> kind >> jam;

  vector<vector<PLL>> G(N);
  REP(i,M){
    LL u, v, t;
    cin >> u >> v >> t;
    --u;
    --v;
    G[u].EB(v, t);
    G[v].EB(u, t);
  }

  const LL BIAS = 1e9;
  VVL dist_pj(N, VL(2, INF));
  {
    // (dist, (vert, flag))
    using S = pair<LL,PII>;
    priority_queue<S,vector<S>,greater<S>> pq;
    pq.push(MP(0, MP(0, 0)));
    dist_pj[0][0] = 0;
    while(!pq.empty()){
      auto s = pq.top();
      pq.pop();

      LL d = s.FF;
      int from = s.SS.FF;
      int flag = s.SS.SS;
      if(dist_pj[from][flag] < d) continue;

      if(flag == 0){
        LL nd = dist_pj[from][flag] + BIAS - pan[from];
        if(dist_pj[from][1] > nd){
          dist_pj[from][1] = nd;
          pq.push(MP(nd, MP(from, 1)));
        }
      }
      for(auto e: G[from]){
        LL nd = d + e.SS;
        int to = e.FF;
        if(dist_pj[to][flag] > nd){
          dist_pj[to][flag] = nd;
          pq.push(MP(nd, MP(to, flag)));
        }  
      }
    }
  }

  VL dist_p(N, INF);
  {
    using S = pair<LL,PII>;
    priority_queue<S,vector<S>,greater<S>> pq;
    pq.push(MP(0, MP(0, 0)));
    dist_p[0] = 0;
    while(!pq.empty()){
      auto s = pq.top();
      pq.pop();

      LL d = s.FF;
      int from = s.SS.FF;
      if(dist_p[from] < d) continue;

      for(auto e: G[from]){
        LL nd = d + e.SS;
        int to = e.FF;
        if(dist_p[to] > nd){
          dist_p[to] = nd;
          pq.push(MP(nd, MP(to, 0)));
        }  
      }
    }
  }

  VL ans(K, -INF);
  REP(i,N){
    int k = kind[i] - 1;
    maxi(ans[k], jam[i] - dist_p[i] - dist_pj[i][1] + BIAS);
  }

  for(auto x : ans) cout << x << endl;

  return 0;
}

