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 int N = 100005, M = 1000005;
using ll = long long;
const ll inf = LLONG_MAX >> 5;
class FlowGraph {
 private:
  int head[N], v[M], nxt[M], ptr, t;
  ll w[M];
  inline void padd_edge(int uu, int vv, ll ww) {
    v[ptr] = vv;
    w[ptr] = ww;
    nxt[ptr] = head[uu];
    head[uu] = ptr++;
  }
  int dis[N], cur[N];
  bool vis[N];
  bool bfs(int s, int t) {
    memset(vis, 0, sizeof(vis));
    memset(dis, -1, sizeof(dis));
    queue<int> q;
    q.push(s);
    dis[s] = 1;
    while (!q.empty()) {
      int nd = q.front();
      q.pop();
      for (int i = head[nd]; i != -1; i = nxt[i])
        if (w[i] && dis[v[i]] == -1) {
          dis[v[i]] = dis[nd] + 1;
          q.push(v[i]);
        }
    }
    return dis[t] != -1;
  }
  ll dfs(int pos, ll flow) {
    if (pos == t) return flow;
    ll tmp = flow;
    for (int &i = cur[pos]; i != -1; i = nxt[i])
      if (w[i] && dis[v[i]] == dis[pos] + 1) {
        ll fl = dfs(v[i], min(flow, w[i]));
        w[i] -= fl;
        w[i ^ 1] += fl;
        flow -= fl;
        if (!flow) break;
      }
    return tmp - flow;
  }

 public:
  FlowGraph() {
    memset(head, -1, sizeof(head));
    memset(nxt, -1, sizeof(nxt));
  }
  inline void add_edge(int uu, int vv, ll ww) {
    padd_edge(uu, vv, ww);
    padd_edge(vv, uu, 0);
  }
  ll dinic(int s, int t) {
    this->t = t;
    ll ret = 0;
    while (bfs(s, t)) {
      memcpy(cur, head, sizeof(cur));
      ret += dfs(s, inf);
    }
    return ret;
  }
};
FlowGraph g;
int n, h, m;
inline int id(int x, int y) { return (x - 1) * (h + 2) + y + 1; }
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> h >> m;
  int s = 0, t = 10000;
  for (int i = 1; i <= n; i++) g.add_edge(s, id(i, 0), inf);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= h + 1; j++)
      if (j != h + 1)
        g.add_edge(id(i, j), id(i, j + 1), h * h - j * j);
      else
        g.add_edge(id(i, j), t, inf);
  int l, r, x, c, lim = id(n, h + 1);
  for (int i = 1; i <= m; i++) {
    cin >> l >> r >> x >> c;
    g.add_edge(lim + i, t, c);
    for (int j = l; j <= r; j++) g.add_edge(id(j, x + 1), lim + i, inf);
  }
  cout << h * h * n - g.dinic(s, t) << endl;
  return 0;
}
