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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

public class Div2553 {

    public static void main(String[] args) throws IOException {

        BufferedReader gi = new BufferedReader(new InputStreamReader(System.in));

        String[] s1 = gi.readLine().split(" ");
        int n = Integer.parseInt(s1[0]);
        int h = Integer.parseInt(s1[1]);
        int m = Integer.parseInt(s1[2]);
        int[] tab1 = new int[n];
        Arrays.fill(tab1, h);
        int[] tab ;
        for (int i = 0; i < m; i++) {
            tab =parseArray(gi);
            for (int j = tab[0]-1; j <= tab[1]-1; j++) {
                tab1[j]=Math.min(tab[2],tab1[j]);
            }
        }
        long result=0;
         PrintWriter pr =new PrintWriter(System.out);
        for (int i = 0; i < tab1.length; i++) {
            result+=(tab1[i]*tab1[i]);
            
            
        }
       
        
        
       
        
      pr.println(result);
        pr.close();

    }

    static int[] parseArray(BufferedReader gi) throws IOException {
        String[] line = gi.readLine().trim().split(" ");
        int[] rez = new int[line.length];
        for (int k = 0; k < line.length; k++) {
            rez[k] = Integer.parseInt(line[k]);
        }
        return rez;
    }
    static boolean[] sieveOfEratosthenes(int n) 
    { 
        // Create a boolean array "prime[0..n]" and initialize 
        // all entries it as true. A value in prime[i] will 
        // finally be false if i is Not a prime, else true. 
        boolean prime[] = new boolean[n+1]; 
        for(int i=0;i<n;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=n; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == true) 
            { 
                // Update all multiples of p 
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 
          return prime ;
        // Print all prime numbers 
      //  int[] tab = new int[n];
         //int x=0;
      /*  for(int i = 2; i <= n; i++) 
        { 
            if(prime[i] == true){
                
            }
                
        } */
    } 

}