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;
const int mn = 52;
int a[mn], b[mn], c[mn], e[mn];
int dp[mn][mn][mn];
int main() {
  int n, h, m, i, j, k, l, ii;
  scanf("%d%d%d", &n, &h, &m);
  for (i = 0; i < m; i++) scanf("%d%d%d%d", a + i, b + i, c + i, e + i);
  for (int d = 0; d < n; d++) {
    for (i = 1; i + d <= n; i++) {
      j = i + d;
      for (k = 0; k <= h; k++) {
        for (l = i; l <= j; l++) {
          int pen = 0;
          for (ii = 0; ii < m; ii++) {
            if (a[ii] >= i && b[ii] <= j && a[ii] <= l && b[ii] >= l &&
                c[ii] < k)
              pen += e[ii];
          }
          dp[i][j][k] =
              max(dp[i][j][k], dp[i][l - 1][k] + dp[l + 1][j][k] + k * k - pen);
        }
        dp[i][j][k + 1] = dp[i][j][k];
      }
    }
  }
  printf("%d", dp[1][n][h]);
}
