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 F>
struct dinic {
  static const F inf = numeric_limits<F>::max();
  int V;
  vector<vector<int>> adj;
  vector<int> dest;
  vector<F> cap;
  dinic(int V = 0) : V(V) { adj.resize(V); }
  void __arc(int u, int v, F c) {
    adj[u].push_back(dest.size());
    dest.push_back(v);
    cap.push_back(c);
  }
  void arc(int u, int v, F c) {
    __arc(u, v, c);
    __arc(v, u, F(0));
  }
  bool bfs(int s, int t, vector<int>& level, vector<F>& flow) const {
    level = vector<int>(V, -1);
    level[s] = 0;
    for (queue<int> q({s}); !q.empty(); q.pop()) {
      int u = q.front();
      for (int e : adj[u]) {
        if (level[dest[e]] == -1 && flow[e] < cap[e]) {
          level[dest[e]] = level[u] + 1;
          q.push(dest[e]);
        }
      }
    }
    return level[t] != -1;
  }
  F augment(int s, int t, vector<int>& level, vector<F>& flow, vector<int>& inx,
            F cur) const {
    if (s == t) return cur;
    for (int e; inx[s] < adj[s].size(); inx[s]++) {
      e = adj[s][inx[s]];
      if (level[dest[e]] != level[s] + 1) continue;
      if (flow[e] == cap[e]) continue;
      F incr =
          augment(dest[e], t, level, flow, inx, min(cur, cap[e] - flow[e]));
      if (incr > F(0)) {
        flow[e] += incr;
        flow[e ^ 1] -= incr;
        return incr;
      }
    }
    return F(0);
  }
  tuple<F, vector<F>> max_flow(int s, int t) const {
    assert(s != t);
    F res(0);
    vector<F> flow(cap.size());
    for (vector<int> level; bfs(s, t, level, flow);) {
      for (vector<int> inx(V, 0);
           F incr = augment(s, t, level, flow, inx, inf);)
        res += incr;
    }
    return make_tuple(res, flow);
  }
};
const int MAXN = 55;
int N, H, M, l[MAXN], r[MAXN], x[MAXN], c[MAXN];
int source() { return 0; }
int sink() { return N + M + N * (H + 1) + 1; }
int house(int i) { return i; }
int restriction(int i) { return N + i; }
int height(int i, int h) { return N + M + (i - 1) * (H + 1) + h; }
int main() {
  scanf("%d %d %d", &N, &H, &M);
  for (int i = 1; i <= M; ++i) {
    scanf("%d %d %d %d", &l[i], &r[i], &x[i], &c[i]);
  }
  dinic<int> mincut(sink() + 1);
  for (int i = 1; i <= N; ++i) {
    mincut.arc(source(), house(i), mincut.inf);
    for (int j = 1; j <= H; ++j) {
      mincut.arc(house(i), height(i, j), j * j - (j - 1) * (j - 1));
    }
  }
  for (int i = 1; i <= M; ++i) {
    for (int j = l[i]; j <= r[i]; ++j) {
      for (int h = x[i] + 1; h <= H; ++h) {
        mincut.arc(height(j, h), restriction(i), mincut.inf);
      }
    }
    mincut.arc(restriction(i), sink(), c[i]);
  }
  tuple<int, vector<int>> flow = mincut.max_flow(source(), sink());
  int answer = N * H * H - get<0>(flow);
  printf("%d\n", answer);
  return 0;
}
