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[51][51][51];
vector<pair<pair<int, int>, pair<int, int> > > ques;
int dp_func(int st, int en, int lo) {
  if (st > en) return 0;
  if (dp[st][en][lo] != -1) return dp[st][en][lo];
  int ret = 0;
  for (int i = st; i <= en; i++) {
    int f = 0, sum = 0;
    for (int j = 0; j <= lo; j++) {
      while (f < ques.size() && ques[f].first.first < j) {
        if (ques[f].second.first >= st && ques[f].second.second <= en &&
            ques[f].second.first <= i && ques[f].second.second >= i)
          sum += ques[f].first.second;
        f++;
      }
      int d = -sum + (j * j) + dp_func(st, i - 1, j) + dp_func(i + 1, en, j);
      ret = max(ret, d);
    }
  }
  return dp[st][en][lo] = ret;
}
int main() {
  int i, j, k, l, m, n, h;
  scanf("%d%d%d", &n, &h, &m);
  for (int i = 1; i <= m; i++) {
    int s;
    scanf("%d%d%d%d", &l, &k, &j, &s);
    ques.push_back(make_pair(make_pair(j, s), make_pair(l, k)));
  }
  sort(ques.begin(), ques.end());
  memset(dp, -1, sizeof dp);
  int ans = dp_func(1, n, h);
  cout << ans << endl;
}
