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, q;
int l[55], r[55], c[55], t[55], ans, custo[55][55][55][55];
int dp[55][55][55];
int solve(int ini, int fim, int vmax) {
  int ans = 0;
  if (fim < ini) return 0;
  if (dp[ini][fim][vmax] != -1) return dp[ini][fim][vmax];
  for (int x = 0; x <= vmax; x++)
    for (int mid = ini; mid <= fim; mid++)
      ans = max(ans, solve(ini, mid - 1, x) + solve(mid + 1, fim, x) + x * x -
                         custo[ini][fim][mid][x]);
  return dp[ini][fim][vmax] = ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> h >> q;
  for (int i = 1; i <= q; i++) cin >> l[i] >> r[i] >> c[i] >> t[i];
  memset(dp, -1, sizeof dp);
  for (int ini = 1; ini <= n; ini++)
    for (int fim = ini; fim <= n; fim++)
      for (int mid = ini; mid <= fim; mid++)
        for (int x = 0; x <= h; x++)
          for (int i = 1; i <= q; i++)
            if (ini <= l[i] and l[i] <= mid and mid <= r[i] and r[i] <= fim and
                x > c[i])
              custo[ini][fim][mid][x] += t[i];
  cout << solve(1, n, h) << "\n";
}
