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 long long mod = 998244353;
const long long INF = mod * mod;
const long double eps = 1e-8;
struct edge {
  int to, cap, rev;
};
vector<edge> G[100000];
bool used[100000];
void add_edge(int from, int to, int cap) {
  G[from].push_back(edge{to, cap, (int)G[to].size()});
  G[to].push_back(edge{from, 0, (int)G[from].size() - 1});
}
int dfs(int v, int t, int f) {
  if (v == t) return f;
  used[v] = true;
  for (int i = 0; i < (int)G[v].size(); i++) {
    edge &e = G[v][i];
    if (!used[e.to] && e.cap > 0) {
      int d = dfs(e.to, t, min(f, e.cap));
      if (d > 0) {
        e.cap -= d;
        G[e.to][e.rev].cap += d;
        return d;
      }
    }
  }
  return 0;
}
int max_flow(int s, int t) {
  int flow = 0;
  for (;;) {
    memset(used, 0, sizeof(used));
    int f = dfs(s, t, mod);
    if (f == 0) return flow;
    flow += f;
  }
}
void solve() {
  int n, h, m;
  cin >> n >> h >> m;
  int ans = h * h * n;
  int s = (h + 1) * n + m;
  int t = s + 1;
  for (int i = 0; i < n; i++) {
    int sta = i * (h + 1);
    for (int j = 1; j <= h; j++) {
      int cost = h * h - (h - j) * (h - j);
      add_edge(sta + j - 1, sta + j, cost);
    }
    add_edge(sta + h, t, mod);
  }
  for (int i = 0; i < m; i++) {
    int l, r, x, c;
    cin >> l >> r >> x >> c;
    l--;
    r--;
    if (x == h) continue;
    int id = (h + 1) * n + i;
    add_edge(s, id, c);
    int xid = h - x - 1;
    for (int j = l; j <= r; j++) {
      add_edge(id, (h + 1) * j + xid, mod);
    }
  }
  ans -= max_flow(s, t);
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
