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 long long ts = 100005;
int dp[52][52][52];
int Left[55], Right[55], x[55], cost[55], Price[55];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n, h, m;
  cin >> n >> h >> m;
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= m; i++) {
    cin >> Left[i] >> Right[i] >> x[i] >> cost[i];
  }
  for (int l = n; l >= 1; l--) {
    for (int r = l; r <= n; r++) {
      for (int k = l; k <= r; k++) {
        memset(Price, 0, sizeof(Price));
        for (int f = 1; f <= m; f++) {
          int L = Left[f];
          int R = Right[f];
          if (L <= k && k <= R && l <= L && R <= r) {
            Price[x[f] + 1] += cost[f];
          }
        }
        for (int f = 1; f <= h; f++) Price[f] += Price[f - 1];
        for (int f = 0; f <= h; f++) {
          dp[l][r][f] =
              max(dp[l][k - 1][f] + dp[k + 1][r][f] - Price[f] + f * f,
                  dp[l][r][f]);
          if (f) {
            dp[l][r][f] = max(dp[l][r][f], dp[l][r][f - 1]);
          }
        }
      }
    }
  }
  cout << dp[1][n][h];
}
