17B. Hierarchy

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B. Hierarchy

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

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: «employeeai 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.

借鉴大神的做法

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=0x3f3f3f3f;
int result(int n,int x[])
{
	int sum=0,kase=1;
	for(int i=1;i<=n;++i)
	{
		if(x[i]==maxn)
		{
			sum-=maxn;
			if(!kase)
			{
				return -1;
			}
			kase=0;
		}
		sum+=x[i];
	}
	return sum;
}
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		int temp,x[1005]={0};
		for(int i=1;i<=n;++i)
		{
			scanf("%d",&temp);
			x[i]=maxn;
		}
		
		int m,a,b,c;
		scanf("%d",&m);
		for(int i=0;i<m;++i)
		{
			scanf("%d%d%d",&a,&b,&c);
			if(x[b]>c)//留下最少花费的 
			{
				x[b]=c;
			}
		}
		printf("%d\n",result(n,x));
	}
	return 0;
}