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;
struct Constr {
  int l, r, x, c;
  bool operator<(const Constr &other) const { return x > other.x; }
};
const int Mx = 59;
long long dp[Mx][Mx][Mx];
long long cost[Mx];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  int N, H, M;
  cin >> N >> H >> M;
  vector<Constr> constraints(M);
  for (Constr &constr : constraints)
    cin >> constr.l >> constr.r >> constr.x >> constr.c;
  sort(constraints.begin(), constraints.end());
  for (int len = 0; len <= N; ++len) {
    for (int i = 1; i + len - 1 <= N; ++i) {
      const int j = i + len - 1;
      vector<Constr> active = constraints;
      active.erase(remove_if(active.begin(), active.end(),
                             [&](const Constr &constr) {
                               return constr.l < i || constr.r > j;
                             }),
                   active.end());
      memset(cost, 0, sizeof cost);
      for (int h = 0; h <= H; ++h) {
        while (!active.empty() && active.back().x < h) {
          for (int k = active.back().l; k <= active.back().r; ++k)
            cost[k] += active.back().c;
          active.pop_back();
        }
        long long &res = dp[i][j][h] = 0;
        if (!len || !h) continue;
        res = dp[i][j][h - 1];
        for (int k = i; k <= j; ++k)
          res = max(res, h * h - cost[k] + dp[i][k - 1][h] + dp[k + 1][j][h]);
        assert(res >= 0);
      }
    }
  }
  cout << dp[1][N][H] << endl;
  assert((int)constraints.size() == M);
  return 0;
}
