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 N = 52;
int n, h, m;
int dp[N][N][N], ssum[N][N][N][N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> h >> m;
  for (int i = 0; i < (m); ++i) {
    int l, r, x, c;
    cin >> l >> r >> x >> c;
    for (int li = 1; li <= l; li++) {
      for (int ri = r; ri <= n; ri++) {
        for (int mx = l; mx <= r; mx++) {
          ssum[li][ri][mx][x + 1] += c;
        }
      }
    }
  }
  for (int li = 1; li <= n; li++) {
    for (int ri = li; ri <= n; ri++) {
      for (int mx = li; mx <= ri; mx++) {
        for (int x = 1; x <= h; x++)
          ssum[li][ri][mx][x] += ssum[li][ri][mx][x - 1];
      }
    }
  }
  for (int k = 1; k <= h; k++) {
    for (int len = 1; len <= n; len++) {
      for (int i = 1; i + len - 1 <= n; i++) {
        int j = i + len - 1;
        dp[i][j][k] = dp[i][j][k - 1];
        for (int mx = i; mx <= j; mx++) {
          dp[i][j][k] =
              max(dp[i][j][k], dp[i][mx - 1][k - 1] + dp[mx + 1][j][k] -
                                   ssum[i][j][mx][k] + k * k);
        }
      }
    }
  }
  cout << dp[1][n][h] << endl;
}
