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 = 52;
int n, h, m;
struct Item {
  int l, r, x, c;
  void read() { scanf("%d%d%d%d", &l, &r, &x, &c); }
} it[N];
int dp[N][N][N], fine[N][N][N][N];
int cac(int l, int r, int h) {
  if (dp[l][r][h] != -1) return dp[l][r][h];
  if (l > r) return dp[l][r][h] = 0;
  dp[l][r][h] = 0;
  for (int f = l; f <= r; f++)
    for (int i = 0; i <= h; i++) {
      dp[l][r][h] =
          max(cac(l, f - 1, i) + cac(f + 1, r, i) + i * i - fine[l][r][f][i],
              dp[l][r][h]);
    }
  return dp[l][r][h];
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%d%d%d", &n, &h, &m);
  for (int i = 1; i <= m; i++) it[i].read();
  for (int l = 1; l <= n; l++)
    for (int r = 1; r <= n; r++) {
      for (int f = 1; f <= r; f++) {
        for (int j = 1; j <= m; j++) {
          for (int i = 0; i <= h; i++) {
            if (it[j].l <= f && f <= it[j].r && it[j].r <= r && it[j].l >= l &&
                it[j].x < i)
              fine[l][r][f][i] += it[j].c;
          }
        }
      }
    }
  cout << cac(1, n, h) << endl;
}
