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;
inline int read() {
  register int data = 0, w = 1;
  char ch = 0;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9')
    data = (data << 1) + (data << 3) + (ch ^ 48), ch = getchar();
  return data * w;
}
int cnt = 1, h[1000005];
struct node {
  int to, next, v;
} w[1000005];
void add(int x, int y, int v) {
  ++cnt;
  w[cnt].to = y;
  w[cnt].next = h[x];
  w[cnt].v = v;
  h[x] = cnt;
  ++cnt;
  w[cnt].to = x;
  w[cnt].next = h[y];
  w[cnt].v = 0;
  h[y] = cnt;
}
int d[1000005], S, T, gap[1000005];
int dfs(int x, int mf) {
  if (x == T) return mf;
  int ret = 0, dlt;
  for (int i = h[x]; i; i = w[i].next) {
    int y = w[i].to;
    if (d[x] == d[y] + 1 && w[i].v) {
      dlt = dfs(y, min(mf, w[i].v));
      ret += dlt;
      w[i].v -= dlt;
      w[i ^ 1].v += dlt;
      mf -= dlt;
      if (!mf || d[S] == T) return ret;
    }
  }
  if (!(--gap[d[x]])) d[S] = T;
  gap[++d[x]]++;
  return ret;
}
int SAP() {
  int ans = 0;
  gap[0] = T;
  while (d[S] < T) ans += dfs(S, 0x3f3f3f3f);
  return ans;
}
int n, lim, m;
int main() {
  n = read(), lim = read(), m = read();
  for (int i = 1; i <= m; i++) {
    int l = read(), r = read(), x = read(), c = read();
    if (x == lim) continue;
    for (int j = l; j <= r; j++) add(j + x * n, n * lim + i, 0x3f3f3f3f);
    add(n * lim + i, n * lim + m + 2, c);
  }
  S = n * lim + m + 1, T = S + 1;
  int ans = lim * lim * n;
  for (int i = 1; i <= n; i++)
    add(S, i, lim * lim), add(i + (lim - 1) * n, T, 0);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= lim - 1; j++)
      add(i + (j - 1) * n, i + j * n, lim * lim - j * j);
  ans -= SAP();
  cout << ans << '\n';
  return 0;
}
