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
// need
#include <iostream>
#include <algorithm>
// data structure
#include <bitset>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#include <complex>
//#include <deque>
#include <valarray>
#include <unordered_map>
#include <unordered_set>
#include <array>
// etc
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <chrono>
#include <random>
#include <numeric>

// input
#define INIT std::ios::sync_with_stdio(false);std::cin.tie(0);
#define VAR(type, ...)type __VA_ARGS__;MACRO_VAR_Scan(__VA_ARGS__);
template<typename T> void MACRO_VAR_Scan(T& t) { std::cin >> t; }
template<typename First, typename...Rest>void MACRO_VAR_Scan(First& first, Rest& ...rest) { std::cin >> first; MACRO_VAR_Scan(rest...); }
#define VEC_ROW(type, n, ...)std::vector<type> __VA_ARGS__;MACRO_VEC_ROW_Init(n, __VA_ARGS__); for(int w_=0; w_<n; ++w_){MACRO_VEC_ROW_Scan(w_, __VA_ARGS__);}
template<typename T> void MACRO_VEC_ROW_Init(int n, T& t) { t.resize(n); }
template<typename First, typename...Rest>void MACRO_VEC_ROW_Init(int n, First& first, Rest& ...rest) { first.resize(n); MACRO_VEC_ROW_Init(n, rest...); }
template<typename T> void MACRO_VEC_ROW_Scan(int p, T& t) { std::cin >> t[p]; }
template<typename First, typename...Rest>void MACRO_VEC_ROW_Scan(int p, First& first, Rest& ...rest) { std::cin >> first[p]; MACRO_VEC_ROW_Scan(p, rest...); }
#define VEC(type, c, n) std::vector<type> c(n);for(auto& i:c)std::cin>>i;
#define MAT(type, c, m, n) std::vector<std::vector<type>> c(m, std::vector<type>(n));for(auto& R:c)for(auto& w:R)std::cin>>w;
// output
#define OUT(dist) std::cout<<(dist);
#define FOUT(n, dist) std::cout<<std::fixed<<std::setprecision(n)<<(dist);
#define SOUT(n, c, dist) std::cout<<std::setw(n)<<std::setfill(c)<<(dist);
#define SP std::cout<<" ";
#define TAB std::cout<<"\t";
#define BR std::cout<<"\n";
#define SPBR(w, n) std::cout<<(w + 1 == n ? '\n' : ' ');
#define ENDL std::cout<<std::endl;
#define FLUSH std::cout<<std::flush;
#define SHOW(dist) {std::cerr << #dist << "\t:" << (dist) << "\n";}
#define SHOWVECTOR(v) {std::cerr << #v << "\t:";for(const auto& xxx : v){std::cerr << xxx << " ";}std::cerr << "\n";}
#define SHOWVECTOR2(v) {std::cerr << #v << "\t:\n";for(const auto& xxx : v){for(const auto& yyy : xxx){std::cerr << yyy << " ";}std::cerr << "\n";}}
#define SHOWQUEUE(a) {auto tmp(a);std::cerr << #a << "\t:";while(!tmp.empty()){std::cerr << tmp.front() << " ";tmp.pop();}std::cerr << "\n";}
#define SHOWSTACK(a) {auto tmp(a);std::cerr << #a << "\t:";while(!tmp.empty()){std::cerr << tmp.top() << " ";tmp.pop();}std::cerr << "\n";}
// utility
#define ALL(a) (a).begin(),(a).end()
#define FOR(w, a, n) for(int w=(a);w<(n);++w)
#define RFOR(w, a, n) for(int w=(n)-1;w>=(a);--w)
#define REP(w, n) for(int w=0;w<int(n);++w)
#define RREP(w, n) for(int w=int(n)-1;w>=0;--w)
#define IN(a, x, b) (a<=x && x<b)
template<class T> inline T CHMAX(T & a, const T b) { return a = (a < b) ? b : a; }
template<class T> inline T CHMIN(T& a, const T b) { return a = (a > b) ? b : a; }
// test
template<class T> using V = std::vector<T>;
template<class T> using VV = V<V<T>>;

template<typename S, typename T>
std::ostream& operator<<(std::ostream& os, std::pair<S, T> p) {
	os << "(" << p.first << ", " << p.second << ")"; return os;
}

#define random_shuffle "USE std::shuffle!";

// type/const
#define int ll
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using PAIR = std::pair<int, int>;
using PAIRLL = std::pair<ll, ll>;
constexpr int INFINT = (1 << 30) - 1;                    // 1.07x10^ 9
constexpr int INFINT_LIM = (1LL << 31) - 1;              // 2.15x10^ 9
constexpr ll INFLL = 1LL << 60;                          // 1.15x10^18
constexpr ll INFLL_LIM = (1LL << 62) - 1 + (1LL << 62);  // 9.22x10^18
constexpr double EPS = 1e-10;
constexpr int MOD = 1000000007;
constexpr double PI = 3.141592653589793238462643383279;
template<class T, size_t N> void FILL(T(&a)[N], const T & val) { for (auto& x : a) x = val; }
template<class ARY, size_t N, size_t M, class T> void FILL(ARY(&a)[N][M], const T & val) { for (auto& b : a) FILL(b, val); }
template<class T> void FILL(std::vector<T> & a, const T & val) { for (auto& x : a) x = val; }
template<class ARY, class T> void FILL(std::vector<std::vector<ARY>> & a, const T & val) { for (auto& b : a) FILL(b, val); }
// ------------>8------------------------------------->8------------



signed main() {
	INIT;

	VAR(int, n, m, k);
	VEC(int, p, n);
	VEC(int, c, n); REP(i, n) --c[i];
	VEC(int, j, n);
	VEC_ROW(int, m, u, v, t);

	struct Edge {
		int to, cost;
		Edge() {}
		Edge(int to, int cost) : to(to), cost(cost) {}
	};
	std::vector<std::vector<Edge>> g(n);

	REP(i, m) {
		--u[i]; --v[i];
		g[u[i]].emplace_back(v[i], t[i]);
		g[v[i]].emplace_back(u[i], t[i]);
	}

	std::vector<int> dist(n);
	std::function<void(int)> Dijkstra = [&](int s) {
		using P = std::pair<int, int>;
		std::priority_queue<P, std::vector<P>, std::greater<P>> pq;
		std::fill(dist.begin(), dist.end(), INFLL);
		dist[s] = 0;
		pq.push(P(0, s));
		while (!pq.empty()) {
			P p = pq.top(); pq.pop();
			int v = p.second;
			if (dist[v] < p.first) continue;
			for (int i = 0; i < g[v].size(); ++i) {
				Edge e = g[v][i];
				if (dist[e.to] > dist[v] + e.cost) {
					dist[e.to] = dist[v] + e.cost;
					pq.push(P(dist[e.to], e.to));
				}
			}
		}
	};
	Dijkstra(0);

	std::priority_queue<PAIR> pq;
	REP(i, n) pq.push(PAIR(p[i] - dist[i], i));

	V<int> ans(k, -INFLL);

	V<int> fixed(n, -INFLL);
	while (!pq.empty()) {
		PAIR p = pq.top(); pq.pop();
		if (fixed[p.second] > -INFLL) continue;
		fixed[p.second] = p.first;
		CHMAX(ans[c[p.second]], p.first + j[p.second] - dist[p.second]);
		for (auto& e : g[p.second]) {
			if (fixed[e.to] <= -INFLL) pq.push(PAIR(p.first - e.cost, e.to));
		}
	}

	REP(i, k) {
		OUT(ans[i])BR;
	}

	return 0;
}
