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 = 100;
long long f[N][N][N];
struct node {
  int l, r, x, c;
} q[N];
int n, m, h;
bool in(int x, int l, int r) { return l <= x && x <= r; }
int main() {
  scanf("%d %d %d", &n, &h, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d %d %d", &q[i].l, &q[i].r, &q[i].x, &q[i].c);
  }
  for (int i = 1; i <= n; i++)
    for (int l = 1; l + i - 1 <= n; l++) {
      int r = l + i - 1;
      for (int hmax = 0; hmax <= h; hmax++) {
        for (int mid = l; mid <= r; mid++) {
          long long v = f[l][mid - 1][hmax] + f[mid + 1][r][hmax] + hmax * hmax;
          for (int j = 1; j <= m; j++)
            if (q[j].x < hmax && in(q[j].l, l, mid) && in(q[j].r, mid, r))
              v -= q[j].c;
          f[l][r][hmax] = max(f[l][r][hmax], v);
        }
        if (hmax) f[l][r][hmax] = max(f[l][r][hmax], f[l][r][hmax - 1]);
      }
    }
  long long ans = 0;
  for (int i = 0; i <= h; i++) ans = max(ans, f[1][n][i]);
  cout << ans;
  return 0;
}
