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 INF = 1e9;
struct Edge {
  int to, cap, flow;
};
struct Graph {
  int n;
  vector<vector<int> > e;
  vector<Edge> edges;
  vector<int> d, c;
  Graph() {}
  Graph(int _n) {
    n = _n;
    e.resize(n);
  }
  void addEdge(int from, int to, int cap) {
    e[from].push_back(edges.size());
    edges.push_back(Edge({to, cap, 0}));
    e[to].push_back(edges.size());
    edges.push_back(Edge({from, 0, 0}));
  }
  bool bfs() {
    d.assign(n, INF);
    c.assign(n, 0);
    vector<int> q(n);
    int qL = 0, qR = 0;
    d[0] = 0;
    q[qR++] = 0;
    while (qL < qR) {
      int v = q[qL++];
      for (int i = 0; i < (int)e[v].size(); i++) {
        Edge cur = edges[e[v][i]];
        if (d[cur.to] > d[v] + 1 && cur.flow < cur.cap) {
          d[cur.to] = d[v] + 1;
          q[qR++] = cur.to;
        }
      }
    }
    return d[n - 1] != INF;
  }
  int dfs(int v, int flow) {
    if (v == n - 1) return flow;
    if (flow == 0) return 0;
    for (int &i = c[v]; i < (int)e[v].size(); i++) {
      Edge cur = edges[e[v][i]];
      if (d[cur.to] != d[v] + 1) continue;
      int pushed = dfs(cur.to, min(flow, cur.cap - cur.flow));
      if (pushed > 0) {
        edges[e[v][i]].flow += pushed;
        edges[e[v][i] ^ 1].flow -= pushed;
        return pushed;
      }
    }
    return 0;
  }
  long long flow() {
    long long flow = 0;
    while (bfs()) {
      while (int pushed = dfs(0, INF)) {
        flow += pushed;
      }
    }
    return flow;
  }
};
int main() {
  int n, m, h;
  while (cin >> n >> h >> m) {
    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];
      x[i]++;
      l[i]--;
      r[i]--;
    }
    Graph gr(1 + n * (h + 1) + m + 1);
    int sz = gr.n;
    long long sum = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 1; j <= h; j++) {
        gr.addEdge(1 + i * (h + 1) + (j - 1), 1 + i * (h + 1) + j, INF);
        gr.addEdge(1 + i * (h + 1) + j, sz - 1, 2 * j - 1);
        sum += 2 * j - 1;
      }
    }
    for (int i = 0; i < m; i++) {
      if (x[i] > h) continue;
      for (int t = l[i]; t <= r[i]; t++) {
        gr.addEdge(1 + n * (h + 1) + i, 1 + t * (h + 1) + x[i], INF);
      }
      gr.addEdge(0, 1 + n * (h + 1) + i, c[i]);
    }
    cout << sum - gr.flow() << endl;
  }
}
