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;
const int maxn = 55, inf = 5e8;
int dp[maxn][maxn][maxn];
int L[maxn], R[maxn], x[maxn], c[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, h, q;
  cin >> n >> h >> q;
  for (int i = 0; i < q; i++) {
    cin >> L[i] >> R[i] >> x[i] >> c[i];
  }
  for (int r = 1; r <= n; r++) {
    for (int l = r; l >= 1; l--) {
      for (int w = 1; w <= h; w++) {
        dp[l][r][w] = dp[l][r][w - 1];
        for (int pos = l; pos <= r; pos++) {
          int num = 0;
          for (int id = 0; id < q; id++) {
            if (l <= L[id] && R[id] <= r && L[id] <= pos && pos <= R[id] &&
                x[id] < w)
              num += c[id];
          }
          dp[l][r][w] = max(dp[l][r][w], dp[l][pos - 1][w] + dp[pos + 1][r][w] -
                                             num + (w * w));
        }
      }
    }
  }
  return cout << dp[1][n][h] << endl, 0;
}
