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
input = sys.stdin.readline

############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))

def getResult(n,q,m,edges,graph):
    maxq = max(q)
    pRoots = [i for i, j in enumerate(q) if j == maxq]
    #employee can be subbordinate for
    canBeSubbordinateFor = []
    count = 0
    for i in range(n):
        canBeSubbordinateForEach = []
        for u,v,c in edges :
            if v == i :
                canBeSubbordinateForEach.append(c)
        canBeSubbordinateFor.append(canBeSubbordinateForEach)
        if len(canBeSubbordinateForEach) == 0 :
            count+=1;
    if count > 1 :
         return  -1

    cost = 0
    for i in range(n):
        if len(canBeSubbordinateFor[i]) > 0:
            cost += min(canBeSubbordinateFor[i])
    return cost


if __name__ == '__main__':
    n = inp()
    q = inlt()
    m = inp()
    #print(m)
    edges = []
    graph = []
    for i in range(n):
        line = []
        for j in range(n):
            line.append(-1)
        graph.append(line)

    for i in range(m):
        ed = inlt()
        ed[0] = ed[0]-1
        ed[1] = ed[1]-1
        edges.append(ed)

    #print(edges)

    for u,v,c in edges:
        if graph[u][v] > c :
            graph[u][v] = c
        else:
            if graph[u][v] == -1 :
                graph[u][v] = c

    #print(graph)

    result = getResult(n,q,m,edges,graph)
    print(result)