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;
constexpr int INF = 1e9 + 7;
constexpr long long INFL = 1e18;
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &V) {
  os << "[";
  for (const auto &x : V) os << x << ", ";
  return os << "]";
}
const int N = 52;
int n, m, h, dp[N][N][N];
struct Restriction {
  int l, r, x, c;
  bool contain(int y, int h) { return l <= y and y <= r and x < h; }
} pen[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> h >> m;
  for (int i = 0; i < m; i++) {
    Restriction &R = pen[i];
    cin >> R.l >> R.r >> R.x >> R.c;
  }
  for (int len = 1; len <= n; len++) {
    for (int a = 1; a + len - 1 <= n; a++) {
      int b = a + len - 1;
      for (int _h = 0; _h <= h; _h++) {
        if (_h > 0) dp[a][b][_h] = dp[a][b][_h - 1];
        for (int piv = a; piv <= b; piv++) {
          int cost = 0;
          for (int i = 0; i < m; i++)
            if (a <= pen[i].l and pen[i].r <= b and pen[i].contain(piv, _h))
              cost += pen[i].c;
          dp[a][b][_h] =
              max(dp[a][b][_h],
                  dp[a][piv - 1][_h] + dp[piv + 1][b][_h] + _h * _h - cost);
        }
      }
    }
  }
  cout << dp[1][n][h] << "\n";
}
