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
import sys
from heapq import heappop as hpp, heappush as hp
def dijkstra(N, s, Edge):
    inf = 10**18
    dist = [inf] * N
    Q = [(0, s)]
    decided = set()
    for _ in range(N):
        while True:
            dn, vn = hpp(Q)
            if vn not in decided:
                decided.add(vn)
                dist[vn] = dn
                break
        for vf, df in Edge[vn]:
            if vf not in decided:
                hp(Q, (dn + df, vf))
    return dist


N, M ,K = map(int, input().split())
P = list(map(int, input().split()))
C = list(map(int, input().split()))
J = list(map(int, input().split()))

Edge1 = [[] for _ in range(N)]
L = []
for _ in range(M):
    u, v, t = map(int, sys.stdin.readline().split())
    u -= 1
    v -= 1
    Edge1[u].append((v, t))
    Edge1[v].append((u, t))
    L.append((u, v, t))

dist = dijkstra(N, 0, Edge1)

Edge2 = [[] for _ in range(3*N)]

for u, v, t in L:
    Edge2[u].append((v, t))
    Edge2[v].append((u, t))
    Edge2[N+u].append((N+v, t))
    Edge2[N+v].append((N+u, t))
inf = 10**17
for i in range(N):
    Edge2[i].append((N+i, inf - P[i]))
    Edge2[N+i].append((2*N+i, inf - J[i] + dist[i]))

dist2 = dijkstra(3*N, 0, Edge2)
Ans = [-10**20]*K
for i in range(N):
    Ans[C[i]-1] = max(Ans[C[i]-1], 2*inf - dist2[2*N+i])
for a in Ans:
    print(a)
