cf 17B

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: «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.

Sample test(s)

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 -

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
#define INF 1<<30
int n,m,tt,f[1010];
int main()
{
      int u,v,w,ans=0;
      scanf("%d",&n);
      for(int i=1;i<=n;i++)
            scanf("%d",&tt),f[i]=INF;
      scanf("%d",&m);
      for(int i=0;i<m;i++)
      {
            scanf("%d%d%d",&u,&v,&w);
            if(f[v]>w)
                  f[v]=w;
      }
      int k=1,flag=1;
      for(int i=1;i<=n;i++)
      {
            if(f[i]==INF&&k)
            {
                  k=0;
                  continue;
            }
            else if(f[i]==INF)
            {
                  flag=0;
                  printf("-1
");
                  break;
            }
            ans+=f[i];
      }
      if(flag)
            printf("%d
",ans);
      return 0;
}