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 N, H, Q;
int dp[55][55][55];
vector<pair<int, pair<int, int>>> a[55], b[55];
int solve(int l, int r, int h) {
  if (l > r || !h) {
    return 0;
  }
  if (dp[l][r][h] != -1) {
    return dp[l][r][h];
  }
  dp[l][r][h] = solve(l, r, h - 1);
  int tot = 0;
  for (int i = l; i <= r; i++) {
    for (auto n : a[i]) {
      if (n.first <= r && n.second.first < h) {
        tot -= n.second.second;
      }
    }
    dp[l][r][h] =
        max(dp[l][r][h], h * h + tot + solve(l, i - 1, h) + solve(i + 1, r, h));
    for (auto n : b[i]) {
      if (n.first >= l && n.second.first < h) {
        tot += n.second.second;
      }
    }
  }
  return dp[l][r][h];
}
int main() {
  cin >> N >> H >> Q;
  for (int i = 0; i <= N; i++) {
    for (int j = 0; j <= N; j++) {
      for (int h = 0; h <= H; h++) {
        dp[i][j][h] = -1;
      }
    }
  }
  while (Q--) {
    int l, r, lim, f;
    cin >> l >> r >> lim >> f;
    a[l].push_back({r, {lim, f}});
    b[r].push_back({l, {lim, f}});
  }
  cout << solve(1, N, H) << endl;
}
