Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employee ai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi. 

Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.

Input

The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.

Output

Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.

Examples

Input

4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5


Output

11


Input

3
1 2 3
2
3 1 2
3 1 3


Output

-1

Note

In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.
import sys, math
from sys import stdin, stdout

rem =10**9+7
inf=10**18
sys.setrecursionlimit(10 ** 6+5)
#from resource import *; setrlimit(RLIMIT_STACK, (RLIM_INFINITY, RLIM_INFINITY))
take = lambda: map(int, stdin.readline().split())
from heapq import heappush,heappop,heapify
from collections import deque
from bisect import *

class UnionFind:
    """
    Class that implements the union-find structure with
    union by rank and find with path compression
    """

    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [0 for x in range(n)]

    def find(self, v):
        if not v == self.parent[v]:
            self.parent[v] = self.find(self.parent[v])
        return self.parent[v]

    def union(self, x, y):
        xRoot = self.find(x)
        yRoot = self.find(y)
        if xRoot == yRoot:
            return
        if self.rank[xRoot] > self.rank[yRoot]:
            self.parent[yRoot] = xRoot
        else:
            self.parent[xRoot] = yRoot
            if self.rank[xRoot] == self.rank[yRoot]:
                self.rank[yRoot] += 1


n=input()
arr=take()
u=UnionFind(n+1)
q=input()
all=[]
for i in range(q):
    a,b,c=take()
    all.append([c,a,b])
all.sort()
ans=0
done=[False for i in range(n+1)]
for c,a,b in all:
    if u.find(a)!=u.find(b) and done[b]==False:
        u.union(a,b)
        done[b]=True
        ans+=c
cnt=0
for i in  range(1,n+1):
    if u.parent[i]==i:
        cnt+=1
if cnt==1:
    print ans
else:
    print -1




