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 N = 55, inF = 1e9 + 100;
int n, h, m, l[N], r[N], x[N], c[N], awi[N][N][N];
int get(int i, int j, int dxb, int h) {
  int res = 0;
  for (int q = 0; q < m; q++)
    if (l[q] <= dxb && l[q] >= i && r[q] <= j && r[q] >= dxb && x[q] < h)
      res += c[q];
  return res;
}
int main() {
  cin >> n >> h >> m;
  for (int i = 0; i < m; i++) cin >> l[i] >> r[i] >> x[i] >> c[i];
  for (int k = 1; k <= h; k++)
    for (int l = 1; l <= n; l++)
      for (int i = 1; i + l - 1 <= n; i++) {
        int j = i + l - 1;
        awi[i][j][k] = awi[i][j][k - 1];
        for (int dxb = i; dxb <= j; dxb++)
          awi[i][j][k] =
              max(awi[i][j][k], awi[i][dxb - 1][k - 1] + awi[dxb + 1][j][k] +
                                    k * k - get(i, j, dxb, k));
      }
  cout << awi[1][n][h];
  return 0;
}
