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 dp[55][55][55];
int cost[55][55][55][55];
int l[55], r[55], x[55], c[55];
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]);
    --l[i], --r[i];
    for (int a = l[i]; ~a; --a)
      for (int b = r[i]; b < n; ++b) {
        for (int j = l[i]; j <= r[i]; ++j) {
          for (int k = x[i] + 1; k <= h; ++k) {
            cost[a][b][j][k] += c[i];
          }
        }
      }
  }
  for (int l = n - 1; ~l; --l) {
    for (int r = l; r < n; ++r) {
      for (int a = 0; a <= h; ++a) {
        for (int i = l; i <= r; ++i) {
          for (int b = 0; b <= a; ++b) {
            dp[l][r][a] = max(dp[l][r][a], (i == l ? 0 : dp[l][i - 1][b]) +
                                               (i == r ? 0 : dp[i + 1][r][b]) +
                                               b * b - cost[l][r][i][b]);
          }
        }
      }
    }
  }
  printf("%d\n", dp[0][n - 1][h]);
  return 0;
}
