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 ll = long long;
using ld = long double;
using namespace std;
const int N = 64;
int n, hei, m;
int ql[N], qr[N], lvl[N], cost[N];
int dp[N][N][N];
int penalty[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> hei >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> ql[i] >> qr[i] >> lvl[i] >> cost[i];
  }
  for (int h = 0; h <= hei; ++h) {
    for (int l = n; l >= 1; --l) {
      for (int r = l; r <= n; ++r) {
        for (int x = l; x <= r; ++x) {
          fill_n(penalty, N, 0);
          for (int i = 1; i <= m; ++i) {
            if (l <= ql[i] && ql[i] <= x && x <= qr[i] && qr[i] <= r) {
              penalty[lvl[i] + 1] += cost[i];
            }
          }
          for (int i = 1; i < N; ++i) {
            penalty[i] += penalty[i - 1];
          }
          for (int mx = 0; mx <= h; ++mx) {
            int cur = mx * mx - penalty[mx];
            dp[h][l][r] =
                max(dp[h][l][r], cur + dp[mx][l][x - 1] + dp[mx][x + 1][r]);
          }
        }
      }
    }
  }
  cout << dp[hei][1][n] << "\n";
  return 0;
}
