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;
int n, h, m;
struct tuhla {
  int l, r;
  int x, c;
};
tuhla a[57];
namespace flow {
int n;
int source;
int sink;
struct tuhla {
  int to;
  int flow, cap;
  int rev;
  tuhla() { to = flow = cap = rev = 0; }
  tuhla(int _to, int _cap, int _rev) {
    to = _to, cap = _cap, flow = 0, rev = _rev;
  }
};
vector<tuhla> v[3007];
int sz[3007];
int lvl[3007];
int pos[3007];
void add_edge(int x, int y, int cap) {
  v[x].push_back(tuhla(y, cap, sz[y]));
  v[y].push_back(tuhla(x, 0, sz[x]));
  ++sz[x], ++sz[y];
}
bool bfs(int st) {
  queue<int> q;
  for (int i = 1; i <= n; ++i) {
    lvl[i] = pos[i] = 0;
  }
  lvl[st] = 1;
  q.push(st);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = 0; i < sz[x]; ++i) {
      if (lvl[v[x][i].to] == 0 && v[x][i].flow != v[x][i].cap) {
        lvl[v[x][i].to] = lvl[x] + 1, q.push(v[x][i].to);
      }
    }
  }
  return (lvl[sink] != 0);
}
int dfs(int vertex, int curr) {
  if (curr == 0 || vertex == sink) {
    return curr;
  }
  for (; pos[vertex] < sz[vertex]; ++pos[vertex]) {
    if (lvl[v[vertex][pos[vertex]].to] == lvl[vertex] + 1) {
      int aux = dfs(
          v[vertex][pos[vertex]].to,
          min(curr, v[vertex][pos[vertex]].cap - v[vertex][pos[vertex]].flow));
      v[vertex][pos[vertex]].flow += aux;
      v[v[vertex][pos[vertex]].to][v[vertex][pos[vertex]].rev].flow -= aux;
      if (aux > 0) {
        return aux;
      }
    }
  }
  return 0;
}
long long maxflow() {
  long long maxflow = 0;
  while (bfs(source) == true) {
    while (1) {
      int flow = dfs(source, 1000000007);
      if (flow == 0) {
        break;
      }
      maxflow += flow;
    }
  }
  return maxflow;
}
}  // namespace flow
void input() {
  scanf("%d%d%d", &n, &h, &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d%d", &a[i].l, &a[i].r, &a[i].x, &a[i].c);
  }
  flow ::n = n * (h + 1) + m + 2;
  flow ::sink = flow ::n;
  flow ::source = flow ::n - 1;
}
void solve() {
  int en = n * (h + 1);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= h; ++j) {
      if (j == 0) {
        flow ::add_edge(flow ::source, i * (h + 1) + j + 1, 1000000007);
      }
      if (j > 0) {
        flow ::add_edge(i * (h + 1) + j + 1, i * (h + 1) + j, 1000000007);
      }
      if (j < h) {
        flow ::add_edge(i * (h + 1) + j + 1, i * (h + 1) + j + 2,
                        h * h - j * j);
      }
      for (int t = 1; t <= m; ++t) {
        if (i + 1 < a[t].l || a[t].r < i + 1) {
          continue;
        }
        if (j > a[t].x) {
          flow ::add_edge(i * (h + 1) + j + 1, en + t, 1000000007);
        }
      }
    }
  }
  for (int i = 1; i <= m; ++i) {
    flow ::add_edge(en + i, flow ::sink, a[i].c);
  }
  printf("%d\n", n * h * h - flow ::maxflow());
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}
