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.
#!/usr/bin/env python
# -*- coding:utf-8 -*-
#

from collections import defaultdict

def readint(): return int(raw_input())
def readfloat(): return float(raw_input())
def readarray(N, foo):
        res = []
        for _ in xrange(N):
                res.append(foo())
        return res
def readlinearray(foo): return map(foo, raw_input().split())

def get_cost_tree(q, applications):
    
    if len(applications) + 2 < len(q):
        return -1
    
    for candidates in applications.values():
        candidates.sort()
    
    cost = 0
    
    for candidates in applications.values():
        cost += candidates[0][0]
    
    return cost

    
if __name__ == '__main__':
    n = readint()
    q = [0] + readlinearray(int)
    m = readint()
    applications = defaultdict(list)
    
    temp = {}
    for _ in xrange(m):
        a, b, c = readlinearray(int)
        last_cost = temp.get((a, b), None)
        if last_cost is not None and c < last_cost:
            applications[b].remove((c, a))
            applications[b].append((c, a))
        else:
            applications[b].append((c, a))
    
    print get_cost_tree(q, applications)
