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 dp[52][52][52];
vector<vector<int> > l(52);
vector<vector<int> > r(52);
vector<vector<int> > c(52);
int main() {
  int n, h, m;
  cin >> n >> h >> m;
  int csum = 0;
  for (int i = 0; i < m; i++) {
    int L, R, X, C;
    cin >> L >> R >> X >> C;
    L--;
    R--;
    l[X].push_back(L);
    r[X].push_back(R);
    c[X].push_back(C);
    csum -= C;
  }
  for (int f = 0; f <= h; f++) {
    for (int i = 0; i < n; i++) {
      for (int j = i; j <= n; j++) {
        if (j > i) {
          dp[f][i][j] = max(dp[f][i][j], dp[f][i][j - 1] + f * f);
        }
        if (f > 0) {
          for (int k = i; k < j; k++) {
            dp[f][i][j] = max(dp[f][i][j], dp[f][i][k] + dp[f - 1][k][j]);
          }
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = i; j <= n; j++) {
        int b = l[f].size();
        for (int a = 0; a < b; a++) {
          if (i <= l[f][a] && r[f][a] < j) {
            dp[f][i][j] += c[f][a];
          }
        }
      }
    }
  }
  cout << csum + dp[h][0][n] << '\n';
}
