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;
long long n, m, h, f[55][55][55];
struct ola {
  int l, r, x, c;
} a[55];
long long cal(int t, int p, int s) {
  if (t > p || s == 0) return 0;
  if (f[t][p][s] != -1) return f[t][p][s];
  f[t][p][s] = cal(t, p, s - 1);
  long long cnt[55];
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= m; i++) {
    if (s > a[i].x && a[i].l >= t && a[i].r <= p) {
      cnt[a[i].l] += a[i].c;
      cnt[a[i].r + 1] -= a[i].c;
    }
  }
  for (int i = t; i <= p; i++) {
    cnt[i] += cnt[i - 1];
    f[t][p][s] =
        max(f[t][p][s], cal(t, i - 1, s) + cal(i + 1, p, s) + s * s - cnt[i]);
  }
  return f[t][p][s];
}
int main() {
  cin >> n >> h >> m;
  memset(f, -1, sizeof(f));
  for (int i = 1; i <= m; i++) {
    cin >> a[i].l >> a[i].r >> a[i].x >> a[i].c;
  }
  cout << cal(1, n, h);
}
