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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<ll, int> P;
const ll INF=1e16;
int main()
{
    int n, m, k; cin>>n>>m>>k;
    ll p[100010], j[100010];
    vector<int> v[100010];
    for(int i=0; i<n; i++) cin>>p[i];
    for(int i=0; i<n; i++){
        int c; cin>>c; c--;
        v[c].push_back(i);
    }
    for(int i=0; i<n; i++) cin>>j[i];
    vector<P> g[100010];
    for(int i=0; i<m; i++){
        int u1, v1; ll t;
        cin>>u1>>v1>>t;
        u1--; v1--;
        g[u1].push_back({t, v1});
        g[v1].push_back({t, u1});
    }
    priority_queue<P, vector<P>, greater<P>> que;
    ll d[2][100010];
    fill(d[0], d[0]+n, INF);
    fill(d[1], d[1]+n, INF);
    ll mx=1e9+7;
    d[0][0]=0;
    que.push({0, 0});
    while(!que.empty()){
        P pr=que.top();
        que.pop();
        int x=pr.second%n;
        int r=pr.second/n;
        if(d[r][x]<pr.first) continue;
        if(r==0){
            if(d[1][x]>d[0][x]+mx-p[x]){
                d[1][x]=d[0][x]+mx-p[x];
                que.push({d[1][x], x+n});
            }
        }
        for(auto q:g[x]){
            int y=q.second;
            if(d[r][y]>d[r][x]+q.first){
                d[r][y]=d[r][x]+q.first;
                que.push({d[r][y], r*n+y});
            }
        }
    }
    for(int i=0; i<k; i++){
        ll ans=-INF;
        for(auto x:v[i]){
            ans=max(ans, -d[1][x]-d[0][x]+mx+j[x]);
        }
        cout<<ans<<endl;
    }
    return 0;
}
