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;
int m, l[52], r[52], a[52], c[52], dp[52][52][52];
int inline Cost(int i, int j, int k, int h) {
  int cost = h * h;
  for (int x = 1; x <= m; ++x) {
    if (i <= l[x] && r[x] <= j && l[x] <= k && k <= r[x] && a[x] < h) {
      cost -= c[x];
    }
  }
  return cost;
}
int main() {
  int n, h;
  cin >> n >> h >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> l[i] >> r[i] >> a[i] >> c[i];
  }
  for (int len = 1; len <= n; ++len) {
    for (int i = 1, j = i + len - 1; j <= n; ++i, ++j) {
      for (int x = 1; x <= h; ++x) {
        int &ans = dp[i][j][x];
        ans = dp[i][j][x - 1];
        for (int k = i; k <= j; ++k) {
          int ansx = Cost(i, j, k, x) + dp[i][k - 1][x] + dp[k + 1][j][x];
          ans = max(ans, ansx);
        }
      }
    }
  }
  cout << dp[1][n][h] << '\n';
  return 0;
}
