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;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
template <typename T, typename... Args>
void read(T &first, Args &...args) {
  read(first);
  read(args...);
}
const int inf = 10000;
struct way {
  int to, flow, rev;
};
const int N = 20010;
struct dinic {
  vector<way> ways[N];
  int n, m, s, t, u, v, f, ans = 0, cuad, cur, cpo[N];
  int depth[N] = {0};
  queue<int> que;
  dinic() {
    s = N - 1;
    t = N - 2;
  }
  bool bfs() {
    memset(depth, 0, sizeof(depth));
    depth[s] = 1;
    que.push(s);
    while (!que.empty()) {
      cur = que.front();
      que.pop();
      for (auto ech : ways[cur])
        if (!depth[ech.to] && ech.flow > 0) {
          depth[ech.to] = depth[cur] + 1;
          que.push(ech.to);
        }
    }
    return depth[t] != 0;
  }
  int dfs(int num, int flow) {
    if (num == t) return flow;
    int cacu = 0;
    for (int &i = cpo[num]; i < ways[num].size(); i++) {
      way cw = ways[num][i];
      if (depth[cw.to] != depth[num] + 1) continue;
      int able = dfs(cw.to, min(flow, cw.flow));
      if (able > 0) {
        flow -= able;
        ways[num][i].flow -= able;
        ways[cw.to][cw.rev].flow += able;
        cacu += able;
      }
      if (flow == 0) break;
    }
    return cacu;
  }
  void addedge(int u, int v, int f) {
    ways[u].push_back((way){v, f, (int)ways[v].size()});
    ways[v].push_back((way){u, 0, (int)ways[u].size() - 1});
  }
  int getans() {
    while (bfs()) {
      memset(cpo, 0, sizeof(cpo));
      while (1) {
        cuad = dfs(s, inf);
        if (cuad == 0) break;
        ans += cuad;
      }
    }
    return ans;
  }
};
int n, h, m, l, r, x, c;
dinic A;
int id(int i, int j) { return i * (h + 10) + j; }
int main() {
  read(n, h, m);
  int ans = n * h * h;
  for (int i = 1; i <= n; i++) {
    A.addedge(A.s, id(i, 0), h * h);
    for (int j = 1; j <= h; j++) {
      A.addedge(id(i, j - 1), id(i, j), h * h - j * j);
    }
  }
  for (int i = 1; i <= m; i++) {
    read(l, r, x, c);
    for (int j = l; j <= r; j++) {
      A.addedge(id(j, x), i + 5000, inf);
    }
    A.addedge(i + 5000, A.t, c);
  }
  cout << ans - A.getans() << endl;
  return 0;
}
