Problem: You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.

In each spot, if a house has height a, you will gain a^2 dollars from it.

The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.

You would like to build houses to maximize your profit. Determine the maximum profit possible.

Input

The first line contains three integers n, h, and m (1 ≤ n,h,m ≤ 50) — the number of spots, the maximum height, and the number of restrictions.

Each of the next m lines contains three integers l_i, r_i, and x_i (1 ≤ l_i ≤ r_i ≤ n, 0 ≤ x_i ≤ h) — left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.

Output

Print a single integer, the maximum profit you can make.

Examples

Input


3 3 3
1 1 1
2 2 3
3 3 2


Output


14


Input


4 10 2
2 3 8
3 4 7


Output


262

Note

In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.

In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.

In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.

In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10).

Solution (in JAVA):

Solution:

import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Problem1162A {


    public static void main(String[] args) throws IOException {
        Problem1162A solution = new Problem1162A();
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        solution.solve(in,out);

        in.closeReader();
        out.close();

    }

    //-----------Solution to the problem--------------------------
    public void solve(InputReader reader,PrintWriter outputWriter) {

        int numberOfSpots,maxHeight,numberOfRestrictions,spots[];
        numberOfSpots = reader.nextInt();
        maxHeight = reader.nextInt();
        numberOfRestrictions = reader.nextInt();

        //initialize the array
        spots = new int[numberOfSpots] ;
        Arrays.fill(spots, maxHeight);

        for (int i = 0; i < numberOfRestrictions; i++) {
            int a = reader.nextInt();
            int b =reader.nextInt();
            int mh = reader.nextInt();
            for (int j = a - 1 ; j < b; j++) {

                if (spots[j] > mh) {
                    spots[j] = mh;
                }

            }
        }


        int sum = 0;
        for (int spot : spots) {
            sum+=spot*spot;
        }

        outputWriter.println(sum);
    }

}



//-----------InputReader class for faster input----------
class InputReader {

    BufferedReader bufferedReader;
    StringTokenizer stringTokenizer;

    public InputReader() {

        bufferedReader = new BufferedReader(new InputStreamReader(System.in));
    }

    /**
     * Splits the Input line by whitespace and returns one word at a time
     * @return next word in the sequence
     */
    public String next() {
        //If current Tokens have been exhausted, or isn't initialized - fetch new line and tokenize it.
        while (stringTokenizer == null || !stringTokenizer.hasMoreElements()) {
            try {
                stringTokenizer = new StringTokenizer(bufferedReader.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return stringTokenizer.nextToken();
    }

    public int nextInt() {

        return Integer.parseInt(next());
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

    public double nextDouble() {
        return Double.parseDouble(next());
    }

    public String nextLine(){
        String str = "";
        try {
            str = bufferedReader.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }

    /**
     * Useful for debugging, it prints to std out in formatted way :- source : message
     * @param source  The origin of the message
     * @param message Description of message
     */
    public static void log(String source,String message){
        System.out.println(source + ":" + message);
    }

    public void closeReader(){
        try {
            bufferedReader.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

}