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 cost[55][55];
int dp[55][55][55];
int main() {
  int n, h, m;
  scanf("%d%d%d", &n, &h, &m);
  struct Zone {
    int l, r, x, c;
  };
  vector<Zone> a(m);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d%d%d", &a[i].l, &a[i].r, &a[i].x, &a[i].c);
  }
  for (int len = 1; len <= n; ++len)
    for (int i = 1; i <= n - len + 1; ++i) {
      memset(cost, 0, sizeof(cost));
      int j = i + len - 1;
      for (int v = 0; v < m; ++v)
        if (a[v].l >= i && a[v].r <= j)
          for (int w = a[v].l; w <= a[v].r; ++w) {
            cost[w][a[v].x] += a[v].c;
          }
      for (int maxh = 1; maxh <= h; ++maxh) {
        int val = 0;
        for (int k = i; k <= j; ++k) {
          int tot = 0;
          for (int l = 0; l <= maxh; ++l) {
            val = max(val, l * l + dp[i][k - 1][l] + dp[k + 1][j][l] - tot);
            tot += cost[k][l];
          }
        }
        dp[i][j][maxh] = val;
      }
    }
  printf("%d\n", dp[1][n][h]);
}
