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;
struct Dinic {
  int nodes, src, dst;
  vector<int> dist, q, work;
  struct edge {
    int to, rev;
    long long f, cap;
  };
  vector<vector<edge>> g;
  Dinic(int x) : nodes(x), g(x), dist(x), q(x), work(x) {}
  void add_edge(int s, int t, long long cap) {
    g[s].push_back((edge){t, int((g[t]).size()), 0, cap});
    g[t].push_back((edge){s, int((g[s]).size()) - 1, 0, 0});
  }
  bool dinic_bfs() {
    fill(dist.begin(), dist.end(), -1);
    dist[src] = 0;
    int qt = 0;
    q[qt++] = src;
    for (int qh = 0; qh < qt; qh++) {
      int u = q[qh];
      for (int i = 0, to = int((g[u]).size()); i < to; ++i) {
        edge &e = g[u][i];
        int v = g[u][i].to;
        if (dist[v] < 0 && e.f < e.cap) dist[v] = dist[u] + 1, q[qt++] = v;
      }
    }
    return dist[dst] >= 0;
  }
  long long dinic_dfs(int u, long long f) {
    if (u == dst) return f;
    for (int &i = work[u]; i < int((g[u]).size()); i++) {
      edge &e = g[u][i];
      if (e.cap <= e.f) continue;
      int v = e.to;
      if (dist[v] == dist[u] + 1) {
        long long df = dinic_dfs(v, min(f, e.cap - e.f));
        if (df > 0) {
          e.f += df;
          g[v][e.rev].f -= df;
          return df;
        }
      }
    }
    return 0;
  }
  long long max_flow(int _src, int _dst) {
    src = _src;
    dst = _dst;
    long long result = 0;
    while (dinic_bfs()) {
      fill(work.begin(), work.end(), 0);
      while (long long delta = dinic_dfs(src, (1LL << 62))) result += delta;
    }
    return result;
  }
};
int n, m, h;
int node(int i, int j) { return i * (h + 1) + j; }
void solve() {
  cin >> n >> h >> m;
  Dinic grafo(n * (h + 1) + m + 2);
  int s = n * (h + 1) + m, t = n * (h + 1) + m + 1, op = n * (h + 1);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < h; j++) {
      grafo.add_edge(node(i, j), node(i, j + 1), h * h - j * j);
      grafo.add_edge(node(i, j + 1), node(i, j), (1LL << 62));
    }
    grafo.add_edge(s, node(i, 0), (1LL << 62));
  }
  for (int i = 0; i < m; i++) {
    int l, r, x, c;
    cin >> l >> r >> x >> c;
    l--;
    r--;
    if (x == h) continue;
    x++;
    int node_op = op + i;
    for (int j = l; j <= r; j++)
      grafo.add_edge(node(j, x), node_op, (1LL << 62));
    grafo.add_edge(node_op, t, c);
  }
  int aux = grafo.max_flow(s, t);
  cout << n * h * h - aux;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}
