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;
struct node {
  int v, fr, c;
} e[110 * 110 * 110];
int n, h, m, tail[110 * 110], cnt = 1, ans;
int S, T, tot, dis[110 * 110], gap[110 * 110];
inline int read() {
  int x = 0, f = 0;
  char c = getchar();
  while (c < '0' || c > '9') f = (c == '-') ? 1 : f, c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return f ? -x : x;
}
void add(int u, int v, int c) {
  e[++cnt] = (node){v, tail[u], c};
  tail[u] = cnt;
  e[++cnt] = (node){u, tail[v], 0};
  tail[v] = cnt;
}
int dfs(int x, int hav) {
  if (x == T) return hav;
  int can = 0, gone = 0;
  for (int p = tail[x], v; p; p = e[p].fr) {
    v = e[p].v;
    if (dis[x] != dis[v] + 1 || !e[p].c) continue;
    can = dfs(v, min(e[p].c, hav - gone));
    gone += can, e[p].c -= can, e[p ^ 1].c += can;
    if (gone == hav) return gone;
  }
  gap[dis[x]]--;
  if (!gap[dis[x]]) dis[S] = T;
  gap[++dis[x]]++;
  return gone;
}
int main() {
  n = read(), h = read(), m = read();
  tot = n * h, S = ++tot, T = ++tot;
  ans = n * h * h;
  for (int i = (1); i <= (n); i++) {
    add(S, i, h * h);
    for (int j = (1); j <= (h - 1); j++)
      add((j - 1) * n + i, j * n + i, h * h - j * j);
  }
  for (int i = (1); i <= (m); i++) {
    int l = read(), r = read(), x = read(), c = read();
    if (x == h) continue;
    int new_ = ++tot;
    add(new_, T, c);
    for (int j = (l); j <= (r); j++) add(x * n + j, new_, 1010580540);
  }
  while (dis[S] < T) ans -= dfs(S, 1010580540);
  printf("%d\n", ans);
  return 0;
}
