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;
void read(int &x) {
  char ch = getchar();
  x = 0;
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + ch - 48, ch = getchar();
}
int g[3000 + 1][3000 + 1], dis[3000 + 1], gap[3000 + 1], be[3000 + 1],
    his[3000 + 1], pre[3000 + 1];
int bz[3000 + 1];
int Next[50 + 1][50 + 1];
int n, h, m, cnt = 0;
int Sap() {
  for (int i = 1; i <= cnt; i++) be[i] = 1;
  gap[0] = cnt;
  int Flow = 0;
  for (int aug = 2147483647, flow = 0, x = 0; dis[0] < cnt;) {
    his[x] = aug, flow = 0;
    for (int i = be[x]; i <= cnt; i++)
      if (g[x][i] && !bz[i] && dis[x] - 1 == dis[i]) {
        flow = 1, pre[i] = x;
        aug = (aug < g[x][i] ? aug : g[x][i]);
        be[x] = i;
        x = i;
        if (x == cnt) {
          Flow += aug;
          while (x) {
            g[x][pre[x]] += aug, g[pre[x]][x] -= aug;
            x = pre[x];
          }
          aug = 2147483647;
        }
        break;
      }
    if (flow) continue;
    int j = 0;
    for (int i = 1; i <= cnt; i++)
      if (g[x][i] && !bz[i] && (dis[i] < dis[j] || !j)) j = i;
    if (j) {
      if (--gap[dis[x]] == 0) break;
      ++gap[dis[x] = dis[j] + 1];
      be[x] = j;
    } else
      bz[x] = 1;
    if (x) x = pre[x], aug = his[x];
  }
  return Flow;
}
int main() {
  read(n), read(h), read(m);
  cnt = m;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= h; j++) Next[i][j] = ++cnt;
  ++cnt;
  for (int i = 1, l, r, x, c; i <= m; i++) {
    read(l), read(r), read(x), read(c);
    if (x < h)
      for (int j = l; j <= r; j++) g[Next[j][x + 1]][i] = 2147483647;
    g[i][cnt] = c;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= h; j++)
      g[Next[i][j - 1]][Next[i][j]] = (h * h) - ((j - 1) * (j - 1));
  for (int i = 1; i <= n; i++) g[0][Next[i][0]] = 2147483647;
  int ans = (h * h) * n - Sap();
  printf("%d\n", ans);
  return 0;
}
