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 n, h, m;
int f[55][55][55];
int pl[55], pr[55], ph[55], pc[55];
int s[55];
int main() {
  cin >> n >> h >> m;
  for (int i = 0; i < m; i++) {
    cin >> pl[i] >> pr[i] >> ph[i] >> pc[i];
    pl[i]--;
  }
  for (int x = 1; x <= n; x++)
    for (int l = 0; l <= n - x; l++) {
      int r = l + x;
      for (int c = l; c < r; c++) {
        memset(s, 0, sizeof s);
        for (int i = 0; i < m; i++)
          if (l <= pl[i] && pr[i] <= r && pl[i] <= c && c < pr[i])
            s[ph[i] + 1] += pc[i];
        for (int i = 0; i < h; i++) s[i + 1] += s[i];
        for (int p = 0; p <= h; p++)
          for (int u = 0; u <= p; u++) {
            f[l][r][p] =
                max(f[l][r][p], f[l][c][u] + f[c + 1][r][u] + u * u - s[u]);
          }
      }
    }
  printf("%d\n", f[0][n][h]);
  return 0;
}
