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;
using LL = long long;
struct Node {
  int l, r, x, c;
};
const int MAXN = 52;
const int MAXM = 52;
const int MAXH = 52;
int n, h, m;
Node x[MAXM];
int f[MAXN][MAXN][MAXH];
bool z[MAXN][MAXN][MAXH];
int DP(int a, int b, int c) {
  if (a > b) return 0;
  if (z[a][b][c]) return f[a][b][c];
  z[a][b][c] = true;
  if (c == 0) return f[a][b][c] = 0;
  for (int mid = a; mid <= b; ++mid) {
    vector<int> g(c + 1);
    for (int i = 0; i < m; ++i) {
      if (x[i].l < a || b < x[i].r) continue;
      if (x[i].l <= mid && mid <= x[i].r && x[i].x <= c) {
        g[x[i].x] += x[i].c;
      }
    }
    int temp = 0;
    for (int i = 0; i <= c; ++i) {
      f[a][b][c] =
          max(f[a][b][c], DP(a, mid - 1, i) + DP(mid + 1, b, i) + i * i - temp);
      temp += g[i];
    }
  }
  return f[a][b][c];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> h >> m;
  for (int i = 0; i < m; ++i) {
    cin >> x[i].l >> x[i].r >> x[i].x >> x[i].c;
    --x[i].l;
    --x[i].r;
  }
  cout << DP(0, n - 1, h) << endl;
  return 0;
}
