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 = 0x3f3f3f3f;
const long long int llINF = (long long)(1e18) + 100;
const int MAXN = 1e5 + 3000;
const int MAXM = 9e5;
int sq(int x) { return x * x; }
int ned, first[MAXN], work[MAXN], dist[MAXN], q[MAXN];
int cap[MAXM], to[MAXM], nxt[MAXM];
void init() {
  memset(first, -1, sizeof first);
  ned = 0;
}
void add(int u, int v, int f) {
  to[ned] = v, cap[ned] = f;
  nxt[ned] = first[u];
  first[u] = ned++;
  to[ned] = u, cap[ned] = 0;
  nxt[ned] = first[v];
  first[v] = ned++;
}
int dfs(int u, int f, int t) {
  if (u == t) return f;
  for (int &e = work[u]; e != -1; e = nxt[e]) {
    int v = to[e];
    if (dist[v] == dist[u] + 1 && cap[e] > 0) {
      int df = dfs(v, min(f, cap[e]), t);
      if (df > 0) {
        cap[e] -= df;
        cap[e ^ 1] += df;
        return df;
      }
    }
  }
  return 0;
}
bool bfs(int s, int t) {
  memset(&dist, -1, sizeof dist);
  dist[s] = 0;
  int st = 0, en = 0;
  q[en++] = s;
  while (en > st) {
    int u = q[st++];
    for (int e = first[u]; e != -1; e = nxt[e]) {
      int v = to[e];
      if (dist[v] < 0 && cap[e] > 0) {
        dist[v] = dist[u] + 1;
        q[en++] = v;
      }
    }
  }
  return dist[t] >= 0;
}
int dinic(int s, int t) {
  int flow = 0, f;
  while (bfs(s, t)) {
    memcpy(work, first, sizeof work);
    while (f = dfs(s, INF, t)) flow += f;
  }
  return flow;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, h;
  cin >> n >> h >> m;
  int l[55], r[55], hgt[55], c[55];
  for (int i = 0; i < m; i++) cin >> l[i] >> r[i] >> hgt[i] >> c[i];
  init();
  int src = 0;
  int snk = 1;
  map<pair<int, int>, int> id;
  int cnt = 2;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < h + 1; j++) id[{i, j}] = cnt++;
  for (int i = 1; i <= n; i++) {
    add(src, id[{i, 0}], INF);
    for (int j = 1; j <= h; j++) {
      add(id[{i, j}], id[{i, j - 1}], INF);
      add(id[{i, j - 1}], id[{i, j}], sq(h) - sq(j - 1));
    }
  }
  for (int k = 0; k < m; k++) {
    add(cnt, snk, c[k]);
    if (hgt[k] < h)
      for (int i = l[k]; i <= r[k]; i++) add(id[{i, hgt[k] + 1}], cnt, INF);
    cnt++;
  }
  cout << (sq(h) * n - dinic(src, snk)) << '\n';
}
