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;
template <typename T1, typename T2>
inline void chkmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chkmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
const string FILENAME = "input";
const int MAXN = 55;
int n, h, m;
int l[MAXN], r[MAXN], x[MAXN], c[MAXN];
int dp[MAXN][MAXN][MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> h >> m;
  for (int i = 0; i < m; i++) {
    cin >> l[i] >> r[i] >> x[i] >> c[i];
    l[i]--, r[i]--;
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i; j <= n; j++) {
      for (int cc = 0; cc <= h; cc++) {
        dp[i][j][cc] = -1e9;
        for (int f = i; f <= j; f++) {
          int sum = 0;
          for (int k = 0; k < m; k++) {
            if (l[k] <= f && f <= r[k] && l[k] >= i && r[k] <= j && x[k] < cc) {
              sum += c[k];
            }
          }
          chkmax(dp[i][j][cc], (f - 1 < i ? 0 : dp[i][f - 1][cc]) +
                                   (f + 1 > j ? 0 : dp[f + 1][j][cc]) +
                                   cc * cc - sum);
        }
        if (cc) {
          chkmax(dp[i][j][cc], dp[i][j][cc - 1]);
        }
      }
    }
  }
  cout << dp[0][n - 1][h] << '\n';
  return 0;
}
