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>
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,fma")
using namespace std;
__inline__ int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
struct Restriction {
  int l, r, x, c;
  bool operator<(const Restriction& ot) const { return x < ot.x; }
};
__inline__ void remax(int& x, int y) { x = max(x, y); }
const int N = 51;
int dp[N][N][N];
int main() {
  const int n = nxt(), h = nxt(), m = nxt();
  vector<Restriction> a(m);
  for (int i = 0; i < m; ++i) {
    cin >> a[i].l >> a[i].r >> a[i].x >> a[i].c;
    a[i].l -= 1;
  }
  sort(a.begin(), a.end());
  for (int len = 1; len <= n; ++len) {
    for (int l = 0; l + len <= n; ++l) {
      const int r = l + len;
      for (int hh = h; hh >= 0; --hh) {
        for (int i = l; i < r; ++i) {
          for (int h1 = 0; h1 <= hh; ++h1) {
            int penalty = 0;
            for (const auto& res : a) {
              if (res.x >= h1) {
                break;
              } else if (i >= res.l && i < res.r && res.l >= l && res.r <= r) {
                penalty += res.c;
              }
            }
            remax(dp[l][r][hh],
                  dp[l][i][h1] + dp[i + 1][r][h1] + h1 * h1 - penalty);
          }
        }
      }
    }
  }
  cout << dp[0][n][h] << "\n";
  return 0;
}
