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;
constexpr int INF = 1e9 + 7;
constexpr long long INFL = 1e18;
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &V) {
  os << "[";
  for (const auto &x : V) os << x << ", ";
  return os << "]";
}
constexpr int maxn = 2505 * 4;
int n, h, m, nxt[maxn], c[maxn * 10], dist[maxn], id, source, sink;
vector<pair<int, int>> G[maxn];
void add_edge(int a, int b, int cap) {
  G[a].emplace_back(b, id);
  c[id++] = cap;
  G[b].emplace_back(a, id);
  c[id++] = 0;
}
bool bfs() {
  fill_n(dist, maxn, INF);
  dist[source] = 0;
  queue<int> Q;
  Q.push(source);
  while (not Q.empty()) {
    auto w = Q.front();
    Q.pop();
    for (auto e : G[w]) {
      if (c[e.second] > 0 and dist[e.first] == INF) {
        dist[e.first] = dist[w] + 1;
        Q.push(e.first);
      }
    }
  }
  return dist[sink] != INF;
}
int flow(int a, int restr) {
  if (a == sink or restr == 0) return restr;
  int ret = 0;
  for (int &i = nxt[a]; i < (int)G[a].size(); i++) {
    auto &e = G[a][i];
    if (c[e.second] == 0 or dist[e.first] != dist[a] + 1) continue;
    auto f = flow(e.first, min(restr, c[e.second]));
    c[e.second] -= f;
    c[e.second ^ 1] += f;
    restr -= f;
    ret += f;
    if (restr == 0) break;
  }
  return ret;
}
int dinic() {
  int ret = 0;
  while (bfs()) {
    fill_n(nxt, maxn, 0);
    ret += flow(source, INF);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> h >> m;
  source = 0, sink = 1;
  auto get_ind = [&](int a, int b) {
    if (b >= h + 1) return sink;
    if (b == 0) return source;
    return (a - 1) * h + b + 1;
  };
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= h; j++) {
      int a = get_ind(i, j), b = get_ind(i, j + 1);
      add_edge(a, b, h * h - j * j);
      add_edge(b, a, INF);
    }
  for (int i = 0; i < m; i++) {
    int l, r, x, cost;
    cin >> l >> r >> x >> cost;
    int node = n * h + i + 3;
    add_edge(node, sink, cost);
    for (int j = l; j <= r; j++) add_edge(get_ind(j, x + 1), node, INF);
  }
  cout << n * h * h - dinic() << "\n";
}
