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;
const int maxn = 100010;
int to[maxn << 1], nxt[maxn << 1], head[maxn], val[maxn << 1], dis[maxn];
int n, tot, cnt, m, H, S, T, ans;
queue<int> q;
void add(int a, int b, int c) {
  to[cnt] = b, val[cnt] = c, nxt[cnt] = head[a], head[a] = cnt++;
  to[cnt] = a, val[cnt] = 0, nxt[cnt] = head[b], head[b] = cnt++;
}
int dfs(int x, int mf) {
  if (x == T) return mf;
  int temp = mf, i, k;
  for (i = head[x]; i != -1; i = nxt[i])
    if (val[i] && dis[to[i]] == dis[x] + 1) {
      k = dfs(to[i], min(temp, val[i]));
      if (!k) dis[to[i]] = -1;
      val[i] -= k, val[i ^ 1] += k, temp -= k;
      if (!temp) break;
    }
  return mf - temp;
}
inline int bfs() {
  int u, i;
  memset(dis, 0, sizeof(dis));
  while (!q.empty()) q.pop();
  dis[S] = 1, q.push(S);
  while (!q.empty()) {
    u = q.front(), q.pop();
    for (i = head[u]; i != -1; i = nxt[i])
      if (!dis[to[i]] && val[i]) {
        dis[to[i]] = dis[u] + 1;
        if (to[i] == T) return 1;
        q.push(to[i]);
      }
  }
  return 0;
}
int main() {
  scanf("%d%d%d", &n, &H, &m);
  int i, j, a, b, c, d;
  S = 0, tot = T = n * H + 1, ans = n * H * H;
  memset(head, -1, sizeof(head));
  for (i = 1; i <= n; i++) {
    add(S, (i - 1) * H + 1, H * H);
    for (j = 1; j < H; j++) {
      add((i - 1) * H + j, (i - 1) * H + j + 1, H * H - j * j);
    }
  }
  for (i = 1; i <= m; i++) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    if (c == H) continue;
    tot++;
    for (j = a; j <= b; j++) add((j - 1) * H + c + 1, tot, 1 << 30);
    add(tot, T, d);
  }
  while (bfs()) ans -= dfs(S, 1 << 30);
  printf("%d", ans);
  return 0;
}
