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.
import java.io.*;
import java.util.*;

import static java.lang.Math.*;

public class Solution {

    BufferedReader in;
    PrintWriter out;
    StringTokenizer st;

    int n;
    int[] q;
    int m;
    final int INF = 1 << 29;

    void solve() throws IOException {
        n = ni();
        q = new int[n];
        for (int i = 0; i < n; ++i)
            q[i] = ni();
        m = ni();
        int[][] vertex = new int[2][n];
        for (int i = 0; i < vertex.length; i++) {
            Arrays.fill(vertex[i], INF);
        }
        for (int i = 0; i < m; ++i) {
            int a = ni() - 1;
            int b = ni() - 1;
            int c = ni();
            if (vertex[0][b] > c) {
                vertex[0][b] = c;
                vertex[1][b] = a;
            }
        }
        int ans = 0;
        int cnt = 0;
        int v = -1;
        for (int i = 0; i < n; ++i) {
            if (vertex[0][i] < INF) {
                ans += vertex[0][i];
            } else {
                ++cnt;
                v = i;
            }
        }
        if (cnt == 1) {
            // was = new boolean[n];
            // if (dfs(v, -1)) {
            out.println(ans);
            return;
            // }
        }
        out.println(-1);
    }

    public Solution() throws IOException {
        Locale.setDefault(Locale.US);
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        in.close();
        out.close();
    }

    String ns() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    int ni() throws IOException {
        return Integer.valueOf(ns());
    }

    long nl() throws IOException {
        return Long.valueOf(ns());
    }

    double nd() throws IOException {
        return Double.valueOf(ns());
    }

    public static void main(String[] args) throws IOException {
        new Solution();
    }
}
