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>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
template <class T = int>
class Dinic {
 public:
  struct Edge {
    Edge(int a, T b) {
      to = a;
      cap = b;
    }
    int to;
    T cap;
  };
  Dinic(int n) {
    edges.resize(n);
    this->n = n;
  }
  T maxFlow(int src, int sink) {
    T ans = 0;
    while (bfs(src, sink)) {
      T flow;
      pt = std::vector<int>(n, 0);
      while (flow = dfs(src, sink)) {
        ans += flow;
      }
    }
    return ans;
  }
  void addEdge(int from, int to, T cap = 1) {
    edges[from].push_back(list.size());
    list.push_back(Edge(to, cap));
    edges[to].push_back(list.size());
    list.push_back(Edge(from, 0));
  }

 private:
  int n;
  std::vector<std::vector<int> > edges;
  std::vector<Edge> list;
  std::vector<int> h, pt;
  T dfs(int on, int sink, T flow = 1e9) {
    if (flow == 0) {
      return 0;
    }
    if (on == sink) {
      return flow;
    }
    for (; pt[on] < edges[on].size(); pt[on]++) {
      int cur = edges[on][pt[on]];
      if (h[on] + 1 != h[list[cur].to]) {
        continue;
      }
      T got = dfs(list[cur].to, sink, std::min(flow, list[cur].cap));
      if (got) {
        list[cur].cap -= got;
        list[cur ^ 1].cap += got;
        return got;
      }
    }
    return 0;
  }
  bool bfs(int src, int sink) {
    h = std::vector<int>(n, n);
    h[src] = 0;
    std::queue<int> q;
    q.push(src);
    while (!q.empty()) {
      int on = q.front();
      q.pop();
      for (auto a : edges[on]) {
        if (list[a].cap == 0) {
          continue;
        }
        int to = list[a].to;
        if (h[to] > h[on] + 1) {
          h[to] = h[on] + 1;
          q.push(to);
        }
      }
    }
    return h[sink] < n;
  }
};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n, h, m;
  std::cin >> n >> h >> m;
  Dinic<int> graph(m + n * h + 2);
  int src = m + n * h;
  int sink = src + 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < h; j++) {
      graph.addEdge(m + i * h + j, sink, 2 * j + 1);
    }
  }
  for (int x = 0; x < m; x++) {
    int l, r, lim, cost;
    std::cin >> l >> r >> lim >> cost;
    l--;
    for (int i = l; i < r; i++) {
      for (int j = lim; j < h; j++) {
        graph.addEdge(x, m + i * h + j, 1e8);
      }
    }
    graph.addEdge(src, x, cost);
  }
  std::cout << h * h * n - graph.maxFlow(src, sink) << '\n';
}
