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 operator
import itertools

class Done(Exception):
    pass

empl_count = int(raw_input())
qualification = [(i+1, int(q)) for i, q in enumerate(raw_input().split())]
apps_count = int(raw_input())
apps = [[i] + map(int, raw_input().split()) for i in range(apps_count)]

result = 0
try:
    qualification = sorted(qualification, key=operator.itemgetter(1))
    for q in qualification[:-1]:
        # find minimal valued contract
        # will raise ValueError if something happens
        contract = min([a for a in apps if a[2]==q[0]], key=operator.itemgetter(3))
        # print 'Contract', contract
        result += contract[3]

    raise Done


    if not apps_count or apps_count < empl_count-1:
        raise ValueError

    if apps_count == 1:
        if empl_count == 1:
            raise ValueError
        elif empl_count == 2 and (apps[0][1]-1) in range(2) and (apps[0][2]-1 in range(2)):
            result = apps[0][3]
            raise Done
        else:
            raise ValueError

    # We have more than 2 apps
    qualification = sorted(qualification, key=operator.itemgetter(1))[::-1]
    top = qualification[0]
    if top[1] == qualification[1][1]:
        # 2 people with the same highest qualification
        raise ValueError

    # choose minimal price from avail contracts
    managers = set([top])
    managers_id = [top[0]]
    workers = set(itertools.takewhile(lambda q: q[1] == qualification[1][1],
                                      qualification[1:]))

    # print 'Apps', apps
    while workers:
        # print 'Workers', workers
        # print 'Manager IDs', managers_id
        for work in workers:
            # find minimal valued contract
            # will raise ValueError if something happens
            contract = min([a for a in apps if (a[2]==work[0] and a[1] in managers_id)], key=operator.itemgetter(3))
            # print 'Contract', contract
            result += contract[3]
            managers.add(work)

        managers_id = [m[0] for m in managers]
        workers = set(itertools.takewhile(lambda q: q[1] == qualification[len(managers_id)][1],
                                          qualification[len(managers_id):]))

except (ValueError, IndexError), e:
    # print e
    result = -1
except Done:
    pass
finally:
    print result

