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;
using ll = long long;
using Pii = pair<int, int>;
using Vi = vector<int>;
void run();
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(18);
  run();
  return 0;
}
int uplg(int n) { return 32 - __builtin_clz(n); }
int uplg(ll n) { return 64 - __builtin_clzll(n); }
struct Banned {
  int begin, end, height, cost;
  bool operator<(const Banned& r) const { return height < r.height; }
};
int dp[51][51][51];
vector<Banned> bans;
int solve(int begin, int end, int height) {
  int& ret = dp[begin][end][height];
  if (ret >= 0) return ret;
  ret = 0;
  for (int i = (begin); i < (end); i++) {
    int cost = 0, j = 0;
    for (int h = (0); h < (height + 1); h++) {
      while (j < int((bans).size())) {
        auto& b = bans[j];
        if (b.height >= h) break;
        if (b.begin >= begin && b.end <= end && b.begin <= i &&
            b.end >= i + 1) {
          cost += b.cost;
        }
        j++;
      }
      ret = max(ret, solve(begin, i, h) + solve(i + 1, end, h) - cost + h * h);
    }
  }
  return ret;
}
void run() {
  int n, h, m;
  cin >> n >> h >> m;
  bans.resize(m);
  for (auto& b : (bans)) {
    cin >> b.begin >> b.end >> b.height >> b.cost;
    b.begin--;
  }
  sort((bans).begin(), (bans).end());
  for (int i = (0); i < (n + 1); i++)
    for (int j = (0); j < (n + 1); j++)
      for (int k = (0); k < (h + 1); k++) dp[i][j][k] = -1;
  cout << solve(0, n, h) << endl;
}
