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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return {i, j};
}
struct printer {
  ~printer() { cerr << endl; }
  template <class c>
  printer& operator,(c x) {
    cerr << boolalpha << x;
    return *this;
  }
  printer& operator,(string x) {
    cerr << x;
    return *this;
  }
  template <class c, class d>
  printer& operator,(pair<c, d> x) {
    return *this, "(", x.first, ", ", x.second, ")";
  }
  template <class... d, template <class...> class c>
  printer& operator,(c<d...> x) {
    return *this, range(begin(x), end(x));
  }
  template <class c>
  printer& operator,(rge<c> x) {
    *this, "[";
    for (auto it = x.b; it != x.e; ++it) *this, (it == x.b ? "" : ", "), *it;
    return *this, "]";
  }
};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int my_rand(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
const int N = 52;
int dp[N][N][N];
int penalty[N][N][N][N];
int solve(int l, int r, int lastH) {
  if (l > r) return 0;
  if (dp[l][r][lastH] != -1) return dp[l][r][lastH];
  int ans = 0;
  for (int m = l; m <= r; m++) {
    for (int h = 0; h <= lastH; h++) {
      ans = max(ans, h * h - penalty[l][r][m][h] + solve(l, m - 1, h) +
                         solve(m + 1, r, h));
    }
  }
  return dp[l][r][lastH] = ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, H, m;
  cin >> n >> H >> m;
  memset(dp, -1, sizeof dp);
  vector<int> L(m), R(m), X(m), C(m);
  for (int i = 0; i < m; i++) cin >> L[i] >> R[i] >> X[i] >> C[i];
  for (int l = 1; l <= n; l++) {
    for (int r = l; r <= n; r++) {
      for (int mid = l; mid <= r; mid++) {
        for (int h = 0; h <= H; h++) {
          for (int i = 0; i < m; i++) {
            if (L[i] >= l && R[i] <= r && mid >= L[i] && mid <= R[i]) {
              if (h > X[i]) {
                penalty[l][r][mid][h] += C[i];
              }
            }
          }
        }
      }
    }
  }
  cout << solve(1, n, H) << endl;
}
