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.
# n = int(input())
#
# arr = [str(i) for i in range(n+1)]+[str(i) for i in range(n-1,-1,-1)]
#
# for i in range(n):
#     print(" "*(2*(n-i))+" ".join(arr[:i+1]+arr[:i][::-1]))
#
# print(" ".join(arr))
#
# for i in range(n-1,-1,-1):
#     if i == 0:
#         print(" "*(2*(n-i))+" ".join(arr[:i+1]+arr[:i][::-1]), end ="")
#     else:
#         print(" " * (2 * (n - i)) + " ".join(arr[:i + 1] + arr[:i][::-1]))


# arr = list("olleh")
# n = input()
# for i in n:
#     if i in arr[-1]:
#         arr.pop()
#     if not arr:
#         print("YES")
#         exit()2
# print("NO")
#


# def setbit(val, i, bit):
#
#     num = 1 << i
#
#     if bit:
#         return val | num
#
#     return val & ~num
#
#
# n,m = list(map(int,input().split()))
# a = int(input(),2)
# b = int(input(),2)
# for i in range(m):
#     arr = input().split()
#
#     if arr[0] == "set_a":
#         a = setbit(a, int(arr[1]), int(arr[2]))
#     elif arr[0] == "set_b":
#         b = setbit(b, int(arr[1]), int(arr[2]))
#     else:
#         try:
#             print(bin(a+b)[2:][-int(arr[1])-1],end="")
#         except:
#             print(0, end="")

from collections import defaultdict
n = int(input())
q = list(map(int, input().split()))

count = [0 for _ in range(n)]

m = int(input())
costs = defaultdict(list)
for _ in range(m):
    a, b, c = list(map(int, input().split()))
    costs[b].append(c)
    count[b-1] = 1
if count.count(0) != 1:
    print(-1)
    exit()

total = 0
for b, cost in costs.items():
    total += min(cost)

print(total)