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 maxn = 52;
int n, h, m, f[maxn][maxn][maxn];
struct node {
  int l, r, x, c;
} a[maxn];
int main() {
  scanf("%d %d %d", &n, &h, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d %d %d", &a[i].l, &a[i].r, &a[i].x, &a[i].c);
  }
  memset(f, -1, sizeof(f));
  function<int(int, int, int)> solve = [&](int l, int r, int h) {
    if (l > r || h < 0) return 0;
    if (~f[l][r][h]) return f[l][r][h];
    int ans = solve(l, r, h - 1);
    for (int i = l; i <= r; i++) {
      int mx = h * h + solve(l, i - 1, h - 1) + solve(i + 1, r, h);
      for (int j = 1; j <= m; j++) {
        if (a[j].l >= l && a[j].l <= i && a[j].r <= r && a[j].r >= i &&
            h > a[j].x)
          mx -= a[j].c;
      }
      ans = max(ans, mx);
    }
    return f[l][r][h] = ans;
  };
  printf("%d\n", solve(1, n, h));
  return 0;
}
