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 mlines 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

47 2 3 141 2 52 4 13 4 11 3 5

output

11

input

31 2 323 1 23 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.

预处理+贪心就能够了，由于根仅仅有一个，所以在输入的时候预处理子节点的最小值贪心就好了=。=

#include
            
             #include
             
              #include
              
               #include
               
                #include
                
                 using namespace std;const int INF=0x3ffffff;int f[1100];int main(){    int n,m;    int temp,u,v,w;    while(cin>>n)    {        for(int i=1;i<=n;i++)        {            cin>>temp;            f[i]=INF;        }        cin>>m;        for(int i=1;i<=m;i++)        {            cin>>u>>v>>w;            if(f[v]>w)//预处理最小值                f[v]=w;        }        int flag=1,k=1;        int ans=0;        for(int i=1;i<=n;i++)//仅仅能有一个INF，即根节点        {            if(f[i]==INF&&k)            {                ans-=INF;                k=0;            }            else if(f[i]==INF)            {                flag=0;                break;            }            ans+=f[i];        }        if(flag)            cout<
                 
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