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, l[55], r[55], x[55], c[55], dp[55][55][55];
bool ins(int a, int b, int i) { return a <= l[i] && r[i] <= b; }
int main() {
  scanf("%d%d%d", &n, &h, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d%d", l + i, r + i, x + i, c + i);
    r[i]++;
  }
  for (int j = 1; j <= n; j++)
    for (int i = 1; i + j <= n + 1; i++)
      for (int k = 1; k <= h; k++) {
        dp[i][j][k] = dp[i][j][k - 1];
        for (int ii = i; ii < i + j; ii++) {
          int cur = k * k;
          if (ii != i) cur += dp[i][ii - i][k];
          if (ii + 1 != i + j) cur += dp[ii + 1][i + j - ii - 1][k];
          for (int jj = 0; jj < m; jj++)
            if (k > x[jj] && ins(i, i + j, jj) && !ins(i, ii, jj) &&
                !ins(ii + 1, i + j, jj))
              cur -= c[jj];
          dp[i][j][k] = max(dp[i][j][k], cur);
        }
      }
  printf("%d\n", dp[1][n][h]);
}
