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 max_n = 51, inf = 1000111222;
struct constraint {
  int l, r, h, c;
  void read() {
    scanf("%d%d%d%d", &l, &r, &h, &c);
    --l;
    --r;
  }
};
int n, mx_h, m, dp[max_n][max_n][max_n];
constraint a[max_n];
int main() {
  scanf("%d%d%d", &n, &mx_h, &m);
  for (int i = 0; i < m; ++i) {
    a[i].read();
  }
  for (int h = 1; h <= mx_h; ++h) {
    for (int len = 1; len <= n; ++len) {
      for (int l = 0; l + len <= n; ++l) {
        int r = l + len - 1;
        dp[h][l][r] = dp[h - 1][l][r];
        for (int pos = l; pos <= r; ++pos) {
          int res = h * h;
          for (int i = 0; i < m; ++i) {
            if (l <= a[i].l && a[i].r <= r && a[i].l <= pos && pos <= a[i].r &&
                a[i].h < h) {
              res -= a[i].c;
            }
          }
          if (l < pos) {
            res += dp[h][l][pos - 1];
          }
          if (pos < r) {
            res += dp[h][pos + 1][r];
          }
          dp[h][l][r] = max(dp[h][l][r], res);
        }
      }
    }
  }
  printf("%d\n", dp[mx_h][0][n - 1]);
  return 0;
}
