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>
inline int read() {
  int res = 0;
  bool bo = 0;
  char c;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    bo = 1;
  else
    res = c - 48;
  while ((c = getchar()) >= '0' && c <= '9')
    res = (res << 3) + (res << 1) + (c - 48);
  return bo ? ~res + 1 : res;
}
template <class T>
inline T Min(const T &a, const T &b) {
  return a < b ? a : b;
}
const int N = 1e6 + 5, INF = 0x3f3f3f3f;
int n, h, m, l[N], r[N], x[N], c[N], ecnt = 1, nxt[N], adj[N], go[N], cap[N],
                                     len, que[N], S, T, lev[N], cur[N];
int which(int x, int y) { return 1 + (x - 1) * h + y; }
void add_edge(int u, int v, int w) {
  nxt[++ecnt] = adj[u];
  adj[u] = ecnt;
  go[ecnt] = v;
  cap[ecnt] = w;
  nxt[++ecnt] = adj[v];
  adj[v] = ecnt;
  go[ecnt] = u;
  cap[ecnt] = 0;
}
bool bfs() {
  for (int i = S; i <= T; i++) lev[i] = -1, cur[i] = adj[i];
  lev[que[len = 1] = S] = 0;
  for (int i = 1; i <= len; i++) {
    int u = que[i];
    for (int e = adj[u], v = go[e]; e; e = nxt[e], v = go[e])
      if (cap[e] && lev[v] == -1) {
        lev[que[++len] = v] = lev[u] + 1;
        if (v == T) return 1;
      }
  }
  return 0;
}
int dinic(int u, int flow) {
  if (u == T) return flow;
  int res = 0, delta;
  for (int &e = cur[u], v = go[e]; e; e = nxt[e], v = go[e])
    if (cap[e] && lev[u] < lev[v]) {
      delta = dinic(v, Min(cap[e], flow - res));
      if (delta) {
        cap[e] -= delta;
        cap[e ^ 1] += delta;
        res += delta;
        if (res == flow) break;
      }
    }
  if (res != flow) lev[u] = -1;
  return res;
}
int maxflow() {
  int res = 0;
  while (bfs()) res += dinic(S, INF);
  return res;
}
int main() {
  n = read();
  h = read();
  m = read();
  for (int i = 1; i <= m; i++)
    l[i] = read(), r[i] = read(), x[i] = read() + 1, c[i] = read();
  S = 1;
  T = n * h + m + 2;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= h; j++) {
      add_edge(S, which(i, j), (j << 1) - 1);
      if (j > 1) add_edge(which(i, j), which(i, j - 1), INF);
    }
  for (int i = 1; i <= m; i++)
    if (x[i] <= h)
      for (int j = l[i]; j <= r[i]; j++)
        add_edge(which(j, x[i]), 1 + n * h + i, INF);
  for (int i = 1; i <= m; i++) add_edge(1 + n * h + i, T, c[i]);
  return std::cout << n * h * h - maxflow() << std::endl, 0;
}
