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 = 55;
int n, H, m, l[N], r[N], x[N], c[N], f[N][N][N], S[N][N];
int main() {
  scanf("%d%d%d", &n, &H, &m);
  for (int i = 1; i <= m; i++) scanf("%d%d%d%d", &l[i], &r[i], &x[i], &c[i]);
  memset(f, 233, sizeof(f));
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= H; j++) f[i + 1][i][j] = 0;
  for (int i = n; i; i--)
    for (int j = i; j <= n; j++) {
      for (int k = i; k <= j; k++)
        for (int l = 0; l <= H; l++) S[k][l] = 0;
      for (int p = 1; p <= m; p++)
        if (i <= l[p] && r[p] <= j)
          for (int q = l[p]; q <= r[p]; q++) S[q][x[p] + 1] -= c[p];
      for (int k = i; k <= j; k++)
        for (int l = 1; l <= H; l++) S[k][l] += S[k][l - 1];
      for (int k = i; k <= j; k++)
        for (int l = 0; l <= H; l++)
          f[i][j][l] = max(f[i][j][l],
                           f[i][k - 1][l] + f[k + 1][j][l] + S[k][l] + l * l);
      for (int l = 1; l <= H; l++) f[i][j][l] = max(f[i][j][l], f[i][j][l - 1]);
    }
  printf("%d ", f[1][n][H]);
}
