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 MAXN = 50 + 10;
const int MAXH = 50 + 10;
int N, M, hlim;
int l[MAXN], r[MAXN], x[MAXN], c[MAXN];
int f[MAXN][MAXN][MAXH] = {0};
int calc(int pl, int pr, int k, int h) {
  int ret = h * h;
  for (int i = 1; i <= M; i++)
    if (pl <= l[i] && l[i] <= k && k <= r[i] && r[i] <= pr && h > x[i])
      ret -= c[i];
  return ret;
}
int getnum() {
  int num = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) num = (num << 3) + (num << 1) + ch - '0', ch = getchar();
  return num;
}
int main() {
  N = getnum(), hlim = getnum(), M = getnum();
  for (int i = 1; i <= M; i++)
    l[i] = getnum(), r[i] = getnum(), x[i] = getnum(), c[i] = getnum();
  for (int len = 1; len <= N; len++)
    for (int i = 1; i <= N - len + 1; i++) {
      int j = i + len - 1;
      for (int h = 0; h <= hlim; h++) {
        f[i][j][h] = f[i][j][h - 1];
        for (int k = i; k <= j; k++)
          f[i][j][h] = max(f[i][j][h],
                           f[i][k - 1][h] + f[k + 1][j][h] + calc(i, j, k, h));
      }
    }
  printf("%d\n", f[1][N][hlim]);
  return 0;
}
