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 MaxFlow {
  struct edge {
    int next, v, cap, flow;
  };
  int n;
  vector<int> start, level, used;
  vector<edge> g;
  MaxFlow(int n) : start(n, -1), level(n, -1), used(n, -1) {}
  void add_edge(int u, int v, int c) {
    g.push_back({start[u], v, c, 0});
    start[u] = g.size() - 1;
    g.push_back({start[v], u, 0, 0});
    start[v] = g.size() - 1;
  }
  bool bfs(int s, int t) {
    queue<int> q;
    fill(level.begin(), level.end(), -1);
    q.push(s);
    level[s] = 0;
    while (q.size()) {
      int x = q.front();
      q.pop();
      for (int i = start[x]; i != -1; i = g[i].next) {
        if (g[i].flow < g[i].cap && level[g[i].v] == -1) {
          level[g[i].v] = level[x] + 1;
          q.push(g[i].v);
        }
      }
    }
    return level[t] != -1;
  }
  int dfs(int x, int t, int f) {
    if (x == t) return f;
    for (int& i = used[x]; i != -1; i = g[i].next) {
      int v = g[i].v;
      if (level[v] == level[x] + 1 && g[i].flow < g[i].cap) {
        int z = dfs(v, t, min(f, g[i].cap - g[i].flow));
        if (z > 0) {
          g[i].flow += z;
          g[i ^ 1].flow -= z;
          return z;
        }
      }
    }
    return 0;
  }
  int maxflow(int s, int t) {
    int f = 0;
    while (bfs(s, t)) {
      used = start;
      while (1) {
        int z = dfs(s, t, 1123123123);
        f += z;
        if (!z) break;
      }
    }
    return f;
  }
};
int n, h, m;
int l[55], r[55], x[55], c[55];
int source() { return n * (h + 1) + m; }
int sink() { return source() + 1; }
int polje(int i, int j) { return i + n * j; }
int kazna(int i) { return n * (h + 1) + i; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n >> h >> m;
  for (int i = 0; i < m; i++) {
    cin >> l[i] >> r[i] >> x[i] >> c[i];
    l[i]--, r[i]--;
  }
  MaxFlow mf(n * (h + 1) + m + 2);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= h; j++) {
      mf.add_edge(source(), polje(i, j), j ? 2 * j - 1 : 0);
    }
  }
  for (int i = 0; i < m; i++) {
    mf.add_edge(kazna(i), sink(), c[i]);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= h; j++) {
      for (int k = 0; k < m; k++) {
        if (l[k] <= i && i <= r[k] && j > x[k]) {
          mf.add_edge(polje(i, j), kazna(k), 123123123);
        }
      }
    }
  }
  int f = mf.maxflow(source(), sink());
  cerr << f << '\n';
  cout << n * h * h - f << '\n';
}
