Problem: Bankopolis is an incredible city in which all the n crossroads are located on a straight line and numbered from 1 to n along it. On each crossroad there is a bank office.

The crossroads are connected with m oriented bicycle lanes (the i-th lane goes from crossroad ui to crossroad vi), the difficulty of each of the lanes is known.

Oleg the bank client wants to gift happiness and joy to the bank employees. He wants to visit exactly k offices, in each of them he wants to gift presents to the employees.

The problem is that Oleg don't want to see the reaction on his gifts, so he can't use a bicycle lane which passes near the office in which he has already presented his gifts (formally, the i-th lane passes near the office on the x-th crossroad if and only if min(ui, vi) < x < max(ui, vi))). Of course, in each of the offices Oleg can present gifts exactly once. Oleg is going to use exactly k - 1 bicycle lane to move between offices. Oleg can start his path from any office and finish it in any office.

Oleg wants to choose such a path among possible ones that the total difficulty of the lanes he will use is minimum possible. Find this minimum possible total difficulty.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 80) — the number of crossroads (and offices) and the number of offices Oleg wants to visit.

The second line contains single integer m (0 ≤ m ≤ 2000) — the number of bicycle lanes in Bankopolis.

The next m lines contain information about the lanes.

The i-th of these lines contains three integers ui, vi and ci (1 ≤ ui, vi ≤ n, 1 ≤ ci ≤ 1000), denoting the crossroads connected by the i-th road and its difficulty.

Output

In the only line print the minimum possible total difficulty of the lanes in a valid path, or -1 if there are no valid paths.

Examples

Input

7 4
4
1 6 2
6 2 2
2 4 2
2 7 1


Output

6


Input

4 3
4
2 1 2
1 3 2
3 4 2
4 1 1


Output

3

Note

In the first example Oleg visiting banks by path 1 → 6 → 2 → 4.

Path 1 → 6 → 2 → 7 with smaller difficulity is incorrect because crossroad 2 → 7 passes near already visited office on the crossroad 6.

In the second example Oleg can visit banks by path 4 → 1 → 3.

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Zayakin Andrey
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Input in = new Input(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        private final int INF = Integer.MAX_VALUE / 2;
        int[][][][] dp;
        Graph g;

        public void solve(int testNumber, Input in, PrintWriter out) {
            int n = in.nextInt();
            int k = in.nextInt();
            g = new Graph(n);
            int m = in.nextInt();
            for (int i = 0; i < m; i++) {
                int u = in.nextInt() - 1;
                int v = in.nextInt() - 1;
                int cost = in.nextInt();
                g.addDirectEdge(u, v, cost);
            }

            dp = new int[n][n][k + 1][2];
            for (int[][][] dp1 : dp)
                for (int[][] dp2 : dp1)
                    for (int[] dp3 : dp2)
                        Arrays.fill(dp3, -1);

            int ans = INF;
            for (int i = 0; i < n; i++) {
                int d1 = go(0, i, k - 1, 1);
                int d2 = go(i, n - 1, k - 1, 0);
                ans = Math.min(ans, d1);
                ans = Math.min(ans, d2);
            }
            out.println(ans >= INF ? -1 : ans);
        }

        private int go(int l, int r, int yet, int align) {
            if (yet == 0)
                return 0;
            if (l >= r)
                return INF;

            int cur = dp[l][r][yet][align];
            if (cur != -1) {
                return cur;
            }
            int ans = INF;
            int source = align == 0 ? l : r;
            for (Edge edge : g.edges(source)) {
                if (!inside(l, r, align, edge.to))
                    continue;
                if (align == 0) {
                    int d1 = go(l + 1, edge.to, yet - 1, 1);
                    int d2 = go(edge.to, r, yet - 1, 0);
                    ans = Math.min(ans, Math.min(d1, d2) + edge.weight);
                }
                if (align == 1) {
                    int d1 = go(edge.to, r - 1, yet - 1, 0);
                    int d2 = go(l, edge.to, yet - 1, 1);
                    ans = Math.min(ans, Math.min(d1, d2) + edge.weight);
                }
            }
            return dp[l][r][yet][align] = ans;
        }

        private boolean inside(int l, int r, int align, int to) {
            if (align == 0) {
                return to > l && to <= r;
            }
            return to >= l && to < r;
        }

    }

    static class Edge {
        public final int to;
        public final int from;
        public final int weight;

        Edge(int from, int to, int weight) {
            this.to = to;
            this.from = from;
            this.weight = weight;
        }

        Edge(int from, int to) {
            this(from, to, 0);
        }

    }

    static class Input {
        private StringTokenizer tokenizer = null;
        private BufferedReader reader;

        public Input(InputStream inputStream) {
            reader = new BufferedReader(new InputStreamReader(inputStream));
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException();
            }
        }

        public String nextToken() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(nextLine());
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }

    }

    static class Graph {
        private final ArrayList<Edge>[] edges;
        public final int vertexCount;

        public Graph(int vertexCount) {
            this.vertexCount = vertexCount;
            edges = new ArrayList[vertexCount];
            for (int i = 0; i < vertexCount; i++) {
                edges[i] = new ArrayList<>(0);
            }
        }

        public Iterable<Edge> edges(int from) {
            return edges[from];
        }

        public void addDirectEdge(int from, int to, int weight) {
            edges[from].add(new Edge(from, to, weight));
        }

    }
}