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 = 500050;
const int inf = 0x3f3f3f3f;
const long long llinf = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const int MOD = 1e9 + 7;
const double PI = acos(-1.0);
clock_t TIME__START, TIME__END;
void program_end() {}
const int MAXN = 3050, MAXM = 20050, INF = 0x3f3f3f3f;
struct dinic {
  struct Edge {
    int to, nxt, flow;
  } e[MAXM];
  int head[MAXN], ecnt = -1, dis[MAXN], cur[MAXN];
  void add_edge(int u, int v, int f) {
    e[++ecnt] = {v, head[u], f};
    head[u] = ecnt;
    e[++ecnt] = {u, head[v], 0};
    head[v] = ecnt;
  }
  int n, m, s, t;
  bool bfs() {
    memset(dis, 0, sizeof(dis));
    queue<int> q;
    dis[s] = 1;
    q.push(s);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int i = head[u]; ~i; i = e[i].nxt) {
        int v = e[i].to;
        if (!dis[v] && e[i].flow > 0) {
          dis[v] = dis[u] + 1;
          q.push(v);
        }
      }
    }
    return dis[t];
  }
  int dfs(int u, int flow) {
    if (u == t) return flow;
    int nowflow = 0;
    for (int &i = cur[u]; ~i; i = e[i].nxt) {
      int v = e[i].to;
      if (dis[v] == dis[u] + 1 && e[i].flow > 0) {
        int tmp = dfs(v, min(flow, e[i].flow));
        e[i].flow -= tmp;
        e[i ^ 1].flow += tmp;
        flow -= tmp;
        nowflow += tmp;
        if (!flow) break;
      }
    }
    return nowflow;
  }
  int solve(int n, int s, int t) {
    this->s = s, this->t = t, this->n = n;
    int ans = 0;
    while (bfs()) {
      memcpy(cur, head, (n + 5) * sizeof(int));
      ans += dfs(s, INF);
    }
    return ans;
  }
  void init() {
    ecnt = -1;
    memset(head, -1, sizeof(head));
  }
} G;
int n, h, m, S, T;
inline void solve() {
  scanf("%d%d%d", &n, &h, &m);
  G.init();
  G.n = 3000;
  S = G.n - 2, T = G.n - 1;
  for (int i = 1; i <= n; ++i) {
    G.add_edge(S, (i - 1) * (h + 1) + 1, h * h);
    for (int j = 1; j <= h; ++j)
      G.add_edge((i - 1) * (h + 1) + j, (i - 1) * (h + 1) + j + 1,
                 h * h - j * j);
  }
  int offs = 2650;
  for (int i = 1; i <= m; ++i) {
    int l, r, x, c;
    scanf("%d%d%d%d", &l, &r, &x, &c);
    if (x == h) continue;
    for (int j = l; j <= r; ++j)
      G.add_edge((j - 1) * (h + 1) + x + 1, offs + i, inf);
    G.add_edge(offs + i, T, c);
  }
  int ans = n * h * h;
  ans -= G.solve(G.n, S, T);
  cout << ans << '\n';
}
int main() {
  TIME__START = clock();
  int Test = 1;
  while (Test--) {
    solve();
  }
  TIME__END = clock();
  program_end();
  return 0;
}
