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;
namespace max_flow {
const int max_v = 55 * 55, inf = 1000111222;
struct edge {
  int to, flow, capacity;
  edge(int to, int flow, int capacity)
      : to(to), flow(flow), capacity(capacity) {}
};
vector<edge> edges;
vector<int> g[max_v];
void add_edge(int u, int v, int capacity) {
  g[u].push_back(edges.size());
  edges.push_back({v, 0, capacity});
  g[v].push_back(edges.size());
  edges.push_back({u, 0, 0});
}
int h[max_v];
bool bfs(int s, int t) {
  for (int i = 0; i <= t; ++i) {
    h[i] = -1;
  }
  h[s] = 0;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int id : g[v]) {
      if (edges[id].flow < edges[id].capacity && h[edges[id].to] == -1) {
        h[edges[id].to] = h[v] + 1;
        q.push(edges[id].to);
      }
    }
  }
  return h[t] != -1;
}
int dfs(int v, int t, int f) {
  if (v == t) {
    return f;
  }
  for (int id : g[v]) {
    if (edges[id].flow < edges[id].capacity && h[v] + 1 == h[edges[id].to]) {
      int res =
          dfs(edges[id].to, t, min(f, edges[id].capacity - edges[id].flow));
      if (res) {
        edges[id].flow += res;
        edges[id ^ 1].flow -= res;
        return res;
      }
    }
  }
  return 0;
}
int dinic(int s, int t) {
  int res = 0;
  while (bfs(s, t)) {
    while (int f = dfs(s, t, inf)) {
      res += f;
    }
  }
  return res;
}
}  // namespace max_flow
const int max_n = 51, inf = 1000111222;
struct constraint {
  int l, r, h, c;
  void read() {
    scanf("%d%d%d%d", &l, &r, &h, &c);
    --l;
    --r;
  }
};
int n, mx_h, m;
constraint a[max_n];
int main() {
  scanf("%d%d%d", &n, &mx_h, &m);
  for (int i = 0; i < m; ++i) {
    a[i].read();
  }
  int s = n * (mx_h + 1) + m;
  int t = s + 1;
  for (int i = 0; i < n; ++i) {
    max_flow::add_edge(s, i * (mx_h + 1), inf);
    for (int j = 0; j < mx_h; ++j) {
      max_flow::add_edge(i * (mx_h + 1) + j, i * (mx_h + 1) + j + 1,
                         mx_h * mx_h - j * j);
    }
    for (int j = 0; j < m; ++j) {
      if (a[j].l <= i && i <= a[j].r && a[j].h < mx_h) {
        max_flow::add_edge(i * (mx_h + 1) + a[j].h + 1, n * (mx_h + 1) + j,
                           inf);
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    max_flow::add_edge(n * (mx_h + 1) + i, t, a[i].c);
  }
  int ans = n * mx_h * mx_h - max_flow::dinic(s, t);
  printf("%d\n", ans);
  return 0;
}
