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 = 55;
const int inf = 1e9;
int n, m, h, cnt, last[N * N], dis[N * N], cur[N * N], s, t;
struct edge {
  int to, next, c;
} e[N * N * N];
std::queue<int> que;
int point(int x, int y) { return (x - 1) * h + y; }
void addedge(int u, int v, int c) {
  e[++cnt].to = v;
  e[cnt].c = c;
  e[cnt].next = last[u];
  last[u] = cnt;
  e[++cnt].to = u;
  e[cnt].c = 0;
  e[cnt].next = last[v];
  last[v] = cnt;
}
bool bfs() {
  for (int i = s; i <= t; i++) dis[i] = 0;
  while (!que.empty()) que.pop();
  dis[s] = 1;
  que.push(s);
  while (!que.empty()) {
    int x = que.front();
    que.pop();
    for (int i = last[x]; i; i = e[i].next)
      if (e[i].c && !dis[e[i].to]) {
        dis[e[i].to] = dis[x] + 1;
        if (e[i].to == t) return 1;
        que.push(e[i].to);
      }
  }
  return 0;
}
int dfs(int x, int maxf) {
  if (x == t || !maxf) return maxf;
  int ret = 0;
  for (int &i = cur[x]; i; i = e[i].next)
    if (e[i].c && dis[x] + 1 == dis[e[i].to]) {
      int f = dfs(e[i].to, std::min(e[i].c, maxf - ret));
      e[i].c -= f;
      e[i ^ 1].c += f;
      ret += f;
      if (maxf == ret) continue;
    }
  return ret;
}
int dinic() {
  int ans = 0;
  while (bfs()) {
    for (int i = s; i <= t; i++) cur[i] = last[i];
    ans += dfs(s, inf);
  }
  return ans;
}
int main() {
  scanf("%d%d%d", &n, &h, &m);
  s = 0;
  t = h * n + m + 1;
  cnt = 1;
  for (int i = 1; i <= n; i++) {
    addedge(s, point(i, 0), h * h);
    for (int j = 0; j < h - 1; j++)
      addedge(point(i, j), point(i, j + 1), h * h - (j + 1) * (j + 1));
  }
  for (int i = 1; i <= m; i++) {
    int l, r, x, c;
    scanf("%d%d%d%d", &l, &r, &x, &c);
    if (x == h) continue;
    for (int j = l; j <= r; j++) addedge(point(j, x), i + h * n, inf);
    addedge(i + h * n, t, c);
  }
  printf("%d\n", h * h * n - dinic());
  return 0;
}
