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 = 1e4;
const int oo = 1e9;
int fr[101], to[101], x[101], c[101];
bool vis[N];
int dp[101][101][101];
int n, m, h;
int calc(int l, int r, int h) {
  if (l > r || h == 0) return 0;
  int &ret = dp[l][r][h];
  if (ret != -1) return ret;
  ret = calc(l, r, h - 1);
  vector<int> sum(n + 3, 0);
  for (int i = 0; i < m; i++) {
    if (fr[i] >= l && to[i] <= r && h > x[i]) {
      sum[fr[i]] += c[i];
      sum[to[i] + 1] -= c[i];
    }
  }
  for (int i = l + 1; i <= r; i++) sum[i] += sum[i - 1];
  for (int i = l; i <= r; i++) {
    ret = max(ret, calc(l, i - 1, h) + calc(i + 1, r, h) + h * h - sum[i]);
  }
  return ret;
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> n >> h >> m;
  for (int i = 0; i < m; i++) {
    cin >> fr[i] >> to[i] >> x[i] >> c[i];
  }
  cout << calc(1, n, h) << endl;
}
