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/python
import sys
n = int(sys.stdin.next().strip())
qual = map(int, sys.stdin.next().strip().split (" "))
m = int(sys.stdin.next().strip())
apps = []
cost = {}
for i in xrange(m):
    app = map(int, sys.stdin.next().strip().split (" "))
    apps.append(app[:])

maxi = max(qual)
mi = 0
for i in xrange(len(qual)):
    if qual[i] == maxi:
        mi = i;
        break

for i in xrange(n):
    cost[i] = []

cost[mi].append(0)

for i in xrange(m):
    cost[apps[i][1] - 1].append(apps[i][2])

mcost = 0
for i in xrange(n):
    if not (len(cost[i]) > 0):
        print -1
        sys.exit(0)
    mcost += min(cost[i])

print mcost
    
    

