Cf 17B hierarchy

Nick's company employedNPeople. Now Nick needs to build a tree hierarchy? Supervisor-surbodinate? Relations in the company (this is to say that each employee, employee t one, has exactly one supervisor). There areMApplications written in the following form :? EmployeeAIIs ready to become a supervisor of employeeBIAt extra costCI?. The qualificationQJOf each employee is known, and for each application the following is true:QAI?>?QBI.

Wocould 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 integerN(1? ≤?N? ≤? (1000)-amount of employees in the company. The following line containsNSpace-separated numbersQJ(0? ≤?QJ? ≤? 106)-The employees 'qualifications. the following line contains numberM(0? ≤?M? ≤? 10000)-amount of stored ed applications. The followingMLines contain the applications themselves, each of them in the form of three space-separated numbers:AI,BIAndCI(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 applicationQAI?>?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.

Preprocessing + greedy can be done, because there is only one root, so it is better to preprocess the minimum greedy value of the subnode during input =. =

# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <limits. h> 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) // minimum preprocessing value f [v] = W;} int flag = 1, k = 1; int ans = 0; for (INT I = 1; I <= N; I ++) // only one INF can exist, that is, the root node {If (F [I] = inf & K) {ANS-= inf; k = 0;} else if (F [I] = inf) {flag = 0; break;} ans + = f [I];} if (FLAG) cout <ans <Endl; else cout <-1 <Endl;} return 0 ;}