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 maxn = 4005, inf = 1e9 + 7;
int d[maxn], v[maxn], cur[maxn], n, m;
int h;
struct edge {
  int u, v;
  long long c, f;
};
vector<int> g[maxn];
vector<edge> e;
void addedge(int x, int y, int c) {
  int id = e.size();
  g[x].push_back(id);
  e.push_back((edge){x, y, c, 0});
  g[y].push_back(++id);
  e.push_back((edge){y, x, 0, 0});
}
bool bfs(int s, int t) {
  memset(v, 0, sizeof(v));
  d[s] = 0;
  v[s] = 1;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int i = q.front();
    q.pop();
    int sz = g[i].size();
    for (int k = 0; k < sz; k++) {
      int id = g[i][k];
      int j = e[id].v;
      if (v[j] || e[id].c - e[id].f == 0) continue;
      d[j] = d[i] + 1;
      q.push(j);
      v[j] = 1;
    }
  }
  return v[t];
}
long long dfs(int i, long long a, int T) {
  if (a == 0 || i == T) return a;
  long long flow = 0, f;
  int sz = g[i].size();
  for (int &k = cur[i]; k < sz; k++) {
    int id = g[i][k];
    int j = e[id].v;
    if (d[j] == d[i] + 1 && (f = dfs(j, min(a, e[id].c - e[id].f), T)) > 0) {
      e[id].f += f;
      e[id ^ 1].f -= f;
      flow += f;
      a -= f;
      if (a == 0) break;
    }
  }
  return flow;
}
long long maxflow(int s, int t) {
  long long flow = 0;
  while (bfs(s, t)) {
    memset(cur, 0, sizeof(cur));
    flow += dfs(s, inf, t);
  }
  return flow;
}
int idx(int x, int y) { return x * (h + 1) + (y + 1); }
int main() {
  cin >> n >> h >> m;
  int ans = h * h * n;
  int s = 0, t = n * (h + 1) + m + 1;
  for (int i = 0; i < n; i++) {
    addedge(s, idx(i, 0), inf);
    for (int j = 0; j < h; j++) {
      addedge(idx(i, j), idx(i, j + 1), h * h - j * j);
    }
  }
  for (int i = 1; i <= m; i++) {
    int l, r, x, c;
    cin >> l >> r >> x >> c;
    for (int j = l; j <= r; j++) {
      if (x < h) addedge(idx(j - 1, x + 1), n * (h + 1) + i, inf);
    }
    addedge(n * (h + 1) + i, t, c);
  }
  cout << ans - maxflow(s, t) << endl;
  return 0;
}
