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 Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = 51;
int dp[N][N][N];
bool done[N][N][N];
int n, m, h;
int l[N], r[N], x[N], c[N];
vector<int> ind;
int get(int i, int j, int h) {
  if (i > j) return 0;
  if (done[i][j][h]) return dp[i][j][h];
  int mx = 0;
  vector<int> indices;
  for (int k = i; k <= j; k++) {
    indices.clear();
    for (auto t : ind)
      if (l[t] <= k && l[t] >= i && r[t] >= k && r[t] <= j) {
        indices.push_back(t);
      }
    int penalty = 0;
    int pos = 0;
    for (int a = 0; a <= h; a++) {
      while (pos < (int)indices.size() && x[indices[pos]] < a)
        penalty += c[indices[pos++]];
      mx = max(mx, a * a + get(i, k - 1, a) + get(k + 1, j, a) - penalty);
    }
  }
  done[i][j][h] = 1;
  return dp[i][j][h] = mx;
}
int main() {
  scanf("%d", &(n));
  scanf("%d", &(h));
  scanf("%d", &(m));
  for (int i = 1; i <= m; i++)
    scanf("%d", &(l[i])), scanf("%d", &(r[i])), scanf("%d", &(x[i])),
        scanf("%d", &(c[i])), ind.push_back(i);
  sort(ind.begin(), ind.end(), [](int i, int j) { return x[i] < x[j]; });
  printf("%d\n", get(1, n, h));
}
