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 can gain a^2 dollars from it.

The city has m zoning restrictions though. The i-th restriction says that if the tallest house from spots l_i to r_i is strictly more than x_i, you must pay a fine of c_i.

You would like to build houses to maximize your profit (sum of dollars gained minus fines). Determine the maximum profit possible.

Input

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

Each of the next m lines contains four integers l_i, r_i, x_i, c_i (1 ≤ l_i ≤ r_i ≤ n, 0 ≤ x_i ≤ h, 1 ≤ c_i ≤ 5 000).

Output

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

Examples

Input


3 3 3
1 1 1 1000
2 2 3 1000
3 3 2 1000


Output


14


Input


4 10 2
2 3 8 76
3 4 7 39


Output


289

Note

In the first example, it's optimal to build houses with heights [1, 3, 2]. We get a gain of 1^2+3^2+2^2 = 14. We don't violate any restrictions, so there are no fees, so the total profit is 14 - 0 = 14.

In the second example, it's optimal to build houses with heights [10, 8, 8, 10]. We get a gain of 10^2+8^2+8^2+10^2 = 328, and we violate the second restriction for a fee of 39, thus the total profit is 328-39 = 289. Note that even though there isn't a restriction on building 1, we must still limit its height to be at most 10.
#include <bits/stdc++.h>
using namespace std;
int D[51][51][51];
struct point {
  int l, r, h, c;
} w[51];
int n, H, m;
int main() {
  int i, j, k;
  scanf("%d%d%d", &n, &H, &m);
  for (i = 1; i <= m; i++) {
    scanf("%d%d%d%d", &w[i].l, &w[i].r, &w[i].h, &w[i].c);
  }
  for (i = 1; i <= H; i++) {
    for (int L = 1; L <= n; L++) {
      for (int b = 1; b <= n - L + 1; b++) {
        int e = b + L - 1;
        D[i][b][e] = D[i - 1][b][e];
        for (k = b; k <= e; k++) {
          int s = D[i][b][k - 1] + D[i][k + 1][e] + i * i;
          for (j = 1; j <= m; j++) {
            if (w[j].l >= b && w[j].r <= e && w[j].l <= k && w[j].r >= k &&
                w[j].h < i)
              s -= w[j].c;
          }
          D[i][b][e] = max(D[i][b][e], s);
        }
      }
    }
  }
  printf("%d\n", D[H][1][n]);
}
