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 N = 60;
struct interval_t {
  int l, r, c;
};
long long dp[N][N][N], cost[N][N][N];
vector<interval_t> intervals[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, h, m;
  cin >> n >> h >> m;
  for (int k = 1; k <= m; ++k) {
    int l, r, x, c;
    cin >> l >> r >> x >> c;
    intervals[x].push_back({l, r, c});
  }
  for (int k = 1; k <= h; ++k) {
    for (auto &cur : intervals[k - 1]) {
      for (int i = 1; i <= cur.l; ++i) {
        for (int j = cur.r; j <= n; ++j) {
          for (int p = cur.l; p <= cur.r; ++p) {
            cost[i][j][p] += cur.c;
          }
        }
      }
    }
    for (int len = 1; len <= n; ++len) {
      for (int i = 1, j = i + len - 1; j <= n; ++i, ++j) {
        dp[i][j][k] = dp[i][j][k - 1];
        for (int p = i; p <= j; ++p) {
          dp[i][j][k] = max(dp[i][j][k], dp[i][p - 1][k] + dp[p + 1][j][k] +
                                             k * k - cost[i][j][p]);
        }
      }
    }
  }
  cout << dp[1][n][h] << endl;
}
