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<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
using namespace std;

typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 998244353;
const ll INF = mod * mod;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef pair<ll, ll> LP;
typedef vector<LP> vec;
typedef vector<string> svec;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-8;

struct edge {
	int to; ll cost;
};
vector<edge> G[1 << 17];

void solve() {
	int n, m, k;
	cin >> n >> m >> k;
	vector<ll> p(n), c(n), v(n);
	rep(i, n)cin >> p[i];
	rep(i, n)cin >> c[i];
	rep(i, n)cin >> v[i];
	rep(i, m) {
		int a, b; ll c; cin >> a >> b >> c; a--; b--;
		G[a].push_back({ b,c });
		G[b].push_back({ a,c });
	}
	vector<ll> dist(n, INF);
	dist[0] = 0;
	priority_queue<LP, vector<LP>, greater<LP>> q;
	q.push({ 0,0 });
	while (!q.empty()) {
		LP p = q.top(); q.pop();
		int id = p.second;
		if (p.first > dist[id])continue;
		rep(j, G[id].size()) {
			int to = G[id][j].to;
			ll nd = p.first + G[id][j].cost;
			if (nd < dist[to]) {
				dist[to] = nd;
				q.push({ nd,to });
			}
		}
	}
	vector<ll> val(n);
	priority_queue<LP> que;
	rep(i, n) {
		val[i] = p[i] - dist[i];
		//cout << i << " " << val[i] << endl;
		que.push({ val[i],i });
	}
	while (!que.empty()) {
		LP p = que.top(); que.pop();
		int id = p.second;
		if (p.first < val[id])continue;
		rep(j, G[id].size()) {
			int to = G[id][j].to;
			ll nd = p.first - G[id][j].cost;
			if (nd > val[to]) {
				val[to] = nd;
				que.push({ nd,to });
			}
		}
	}
	//cout << val[0] << endl;
	vector<ll> ans(k, -INF);
	rep(i, n) {
		ans[c[i] - 1] = max(ans[c[i]-1],v[i] - dist[i] + val[i]);
	}
	rep(i, k) {
		cout << ans[i] << endl;
	}
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	//cout << fixed << setprecision(10);
	//init();
	solve();
	//stop
	return 0;
}


