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;
int gi() {
  int x = 0, w = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') w = 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return w ? x : -x;
}
const int N = 3005;
const int inf = 1 << 30;
struct edge {
  int to, nxt, w;
} a[N << 3];
int n, h, m, S, T, head[N], cnt = 1, dep[N], q[N], hd, tl, cur[N];
void link(int u, int v, int w) {
  a[++cnt] = (edge){v, head[u], w};
  head[u] = cnt;
  a[++cnt] = (edge){u, head[v], 0};
  head[v] = cnt;
}
bool bfs() {
  memset(dep, 0, sizeof(dep));
  dep[q[hd = tl = 1] = S] = 1;
  while (hd <= tl)
    for (int u = q[hd++], e = head[u]; e; e = a[e].nxt)
      if (a[e].w && !dep[a[e].to]) dep[q[++tl] = a[e].to] = dep[u] + 1;
  return dep[T];
}
int dfs(int u, int f) {
  if (u == T) return f;
  int res = 0;
  for (int &e = cur[u]; e; e = a[e].nxt)
    if (a[e].w && dep[a[e].to] == dep[u] + 1) {
      int d = dfs(a[e].to, min(a[e].w, f));
      a[e].w -= d;
      a[e ^ 1].w += d;
      f -= d;
      res += d;
      if (!f) break;
    }
  if (!res) dep[u] = 0;
  return res;
}
int dinic() {
  int res = 0;
  while (bfs()) {
    memcpy(cur, head, sizeof(cur));
    res += dfs(S, inf);
  }
  return res;
}
int main() {
  n = gi();
  h = gi();
  m = gi();
  S = n * h + 1;
  T = S + 1;
  for (int i = 1; i <= n; ++i) {
    link(S, (i - 1) * h + 1, 1000000);
    for (int j = 1; j < h; ++j)
      link((i - 1) * h + j, (i - 1) * h + j + 1, 1000000 - j * j);
    link(i * h, T, 1000000 - h * h);
  }
  for (int i = 1; i <= m; ++i) {
    int l = gi(), r = gi(), x = gi(), c = gi();
    if (x == h) continue;
    ++x;
    link(T + i, T, c);
    for (int j = l; j <= r; ++j) link((j - 1) * h + x, T + i, inf);
  }
  printf("%d\n", 1000000 * n - dinic());
  return 0;
}
