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 = 60, inf = 1e9 + 42;
struct Edge {
  int u, v;
  long long cap, flow;
  Edge() {}
  Edge(int u, int v, long long cap) : u(u), v(v), cap(cap), flow(0) {}
};
struct Dinic {
  int N;
  vector<Edge> E;
  vector<vector<int>> g;
  vector<int> d, pt;
  Dinic(int N) : N(N), E(0), g(N), d(N), pt(N) {}
  void AddEdge(int u, int v, long long cap) {
    if (u != v) {
      E.emplace_back(Edge(u, v, cap));
      g[u].emplace_back(E.size() - 1);
      E.emplace_back(Edge(v, u, 0));
      g[v].emplace_back(E.size() - 1);
    }
  }
  bool BFS(int S, int T) {
    queue<int> q({S});
    fill(d.begin(), d.end(), N + 1);
    d[S] = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      if (u == T) break;
      for (int k : g[u]) {
        Edge &e = E[k];
        if (e.flow < e.cap && d[e.v] > d[e.u] + 1) {
          d[e.v] = d[e.u] + 1;
          q.emplace(e.v);
        }
      }
    }
    return d[T] != N + 1;
  }
  long long DFS(int u, int T, long long flow = -1) {
    if (u == T || flow == 0) return flow;
    for (int &i = pt[u]; i < g[u].size(); ++i) {
      Edge &e = E[g[u][i]];
      Edge &oe = E[g[u][i] ^ 1];
      if (d[e.v] == d[e.u] + 1) {
        long long amt = e.cap - e.flow;
        if (flow != -1 && amt > flow) amt = flow;
        if (long long pushed = DFS(e.v, T, amt)) {
          e.flow += pushed;
          oe.flow -= pushed;
          return pushed;
        }
      }
    }
    return 0;
  }
  long long MaxFlow(int S, int T) {
    long long total = 0;
    while (BFS(S, T)) {
      fill(pt.begin(), pt.end(), 0);
      while (long long flow = DFS(S, T)) total += flow;
    }
    return total;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, h, m;
  cin >> n >> h >> m;
  Dinic dinic(n * h + m + 2);
  int source = 0, sink = n * h + m + 1;
  for (int i = 0; i < n; ++i) {
    dinic.AddEdge(source, i * h + 1, h * h);
    for (int j = 1; j < h; ++j) {
      dinic.AddEdge(i * h + j, i * h + j + 1, h * h - j * j);
    }
  }
  for (int k = 1; k <= m; ++k) {
    int l, r, x, c;
    cin >> l >> r >> x >> c;
    --l;
    --r;
    if (x == h) {
      continue;
    }
    for (int i = l; i <= r; ++i) {
      dinic.AddEdge(i * h + x + 1, n * h + k, inf);
    }
    dinic.AddEdge(n * h + k, sink, c);
  }
  cout << n * h * h - dinic.MaxFlow(source, sink) << endl;
}
