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;
vector<pair<pair<int, int>, pair<int, int> > > vc;
int con[55][55][55];
int DP[55][55][55], big_dp[55][55][55];
int trans[55][55], sub[55];
int N, H, M;
int main() {
  cin >> N >> H >> M;
  int x, y, z, a;
  for (int i = 0; i < M; i++) {
    cin >> x >> y >> z >> a;
    con[x - 1][y - 1][z] += a;
    vc.push_back(pair<pair<int, int>, pair<int, int> >(
        pair<int, int>(x - 1, y - 1), pair<int, int>(z, a)));
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j <= H; j++) {
      DP[i][i][j] = j * j;
      for (int k = 0; k < j; k++) DP[i][i][j] -= con[i][i][k];
      DP[i][i][j] = max(0, DP[i][i][j]);
    }
    big_dp[i][i][0] = DP[i][i][0];
    for (int k = 1; k <= H; k++)
      big_dp[i][i][k] = max(DP[i][i][k], big_dp[i][i][k - 1]);
  }
  for (int len = 2; len <= N; len++) {
    for (int i = 0; i < N; i++) {
      int j = (i + len) - 1;
      if (j >= N) break;
      for (int k = 0; k <= H; k++)
        for (int l = 0; l < N; l++) {
          trans[l][k] = 0;
          sub[l] = 0;
        }
      for (int k = i; k < j; k++) {
        for (int l = 0; l < M; l++) {
          if (vc[l].first.first >= i && vc[l].first.first <= k &&
              vc[l].first.second >= k + 1 && vc[l].first.second <= j)
            trans[k][vc[l].second.first + 1] += vc[l].second.second;
        }
      }
      for (int k = 0; k <= H; k++) {
        for (int split = i; split < j; split++) {
          sub[split] += trans[split][k];
          DP[i][j][k] =
              max(DP[i][j][k], max(DP[i][split][k] + big_dp[split + 1][j][k],
                                   big_dp[i][split][k] + DP[split + 1][j][k]) -
                                   sub[split]);
        }
      }
      big_dp[i][j][0] = max(0, DP[i][j][0]);
      for (int k = 1; k <= H; k++)
        big_dp[i][j][k] = max(DP[i][j][k], big_dp[i][j][k - 1]);
    }
  }
  cout << big_dp[0][N - 1][H] << endl;
  return (0);
}
