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;
int n, h, m, f[55][55][55], c[55][55][55];
struct P {
  int l, r, x, c;
} p[55];
inline int max(int a, int b, int c) {
  if (a < b) a = b;
  if (a < c) a = c;
  return a;
}
int dfs(int l, int r, int w) {
  if (l > r && w == 0) return 0;
  if (f[l][r][w] != 0) return f[l][r][w];
  for (int i = l; i <= r; i++) {
    int mas = -1e8;
    for (int j = 1; j <= h; j++) c[l][r][j] = 0;
    for (int j = 1; j <= m; j++)
      if (l <= p[j].l && p[j].r <= r && p[j].l <= i && i <= p[j].r)
        c[l][r][p[j].x + 1] += p[j].c;
    for (int j = 1; j <= w; j++) {
      c[l][r][j] += c[l][r][j - 1];
      f[l][r][j] =
          max(f[l][r][j], f[l][r][j - 1],
              dfs(l, i - 1, j) + dfs(i + 1, r, j) + j * j - c[l][r][j]);
    }
  }
  return f[l][r][w];
}
int main() {
  scanf("%d%d%d", &n, &h, &m);
  for (int i = 1; i <= m; i++)
    scanf("%d%d%d%d", &p[i].l, &p[i].r, &p[i].x, &p[i].c);
  printf("%d", dfs(1, n, h));
  return 0;
}
