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>
const int N = 50 * 50 * 2;
const int INF = 1e9;
int to[N << 1], edge, last[N], Next[N << 1], w[N << 1], v[N << 1], dep[N];
int n, m, h, l, r, S, T, vis[N], ans, hh, vv;
std::queue<int> q;
int calh(int x, int y) {
  if (y > h) return T;
  if (y == 0) return S;
  return (x - 1) * h + y;
}
int calr(int x) { return h * n + x; }
void add(int x, int y, int ww) {
  to[++edge] = y;
  Next[edge] = last[x];
  last[x] = edge;
  w[edge] = ww;
}
void Add(int x, int y, int ww) {
  add(x, y, ww);
  add(y, x, 0);
}
bool BFS() {
  memset(dep, 0, sizeof(dep));
  dep[S] = 1;
  q.push(S);
  while (q.size()) {
    int x = q.front();
    q.pop();
    for (int i = last[x]; i; i = Next[i]) {
      int u = to[i];
      if (dep[u] || !w[i]) continue;
      dep[u] = dep[x] + 1;
      q.push(u);
    }
  }
  return dep[T];
}
int c(int x) {
  if (x & 1) return x + 1;
  return x - 1;
}
int DFS(int x, int now) {
  if (x == T) return now;
  int out = 0;
  for (int i = last[x]; i; i = Next[i]) {
    int u = to[i];
    if (w[i] && dep[u] > dep[x]) {
      int t = DFS(u, std::min(now, w[i]));
      w[i] -= t;
      w[c(i)] += t;
      now -= t;
      out += t;
    }
  }
  if (out == 0) dep[x] = 0;
  return out;
}
int main() {
  scanf("%d%d%d", &n, &h, &m);
  ans = n * h * h;
  S = n * h + m + 1, T = S + 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= h; j++) Add(calh(i, j), calh(i, j + 1), h * h - j * j);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d%d", &l, &r, &hh, &vv);
    if (h == hh) continue;
    Add(calr(i), T, vv);
    for (int j = l; j <= r; j++) Add(calh(j, hh + 1), calr(i), INF);
  }
  while (BFS())
    while (int di = DFS(S, INF)) ans -= di;
  printf("%d\n", ans);
}
