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.
'''input
4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5
'''

def root(node):
	while parent[node]!=node:
		parent[node]=parent[parent[node]] # Path Compression
		node = parent[node]
	return node

def union(a,b):
	root_a = root(a)
	root_b = root(b)
	if parent.count(root_a)>parent.count(root_b): # Union by rank
		parent[root_b] = root_a
	else:
		parent[root_a] = root_b


def find(a,b):
	if root(a)==root(b):
		return True
	else:
		return False
n=int(input())
l=list(map(int,input().strip().split(' ')))
q=int(input())
g=[] # adjacency list
for i in range(q):
	u,v,w=map(int,input().strip().split(' '))
	g.append([u-1,v-1,w]) # [src,dest,weight]
parent = [i for i in range(n)]
g.sort(key=lambda x : (x[2]))
# print(g)
mst=[]
e=0
i=0
assigned=[False]*n
while e<n-1 and i<q:
	u=g[i][0]
	v=g[i][1]
	# print(u,root(u),v,root(v))
	if assigned[v]==False and find(u,v)==False:
		assigned[v]=True
		mst.append(g[i])
		union(u,v)
		e+=1
	i+=1
if assigned.count(False)>1:
	print(-1)
else:
	print(sum(x[2] for x in mst))