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 = 55;
const int M = N * N;
const int inf = 1e9 + 7;
struct Edge {
  int u, v;
  long long c;
  Edge() {}
  Edge(int a, int b, long long d) : u(a), v(b), c(d) {}
};
vector<Edge> edges;
vector<int> G[M];
void AddEdge(int u, int v, long long c) {
  G[u].push_back(edges.size());
  edges.push_back(Edge(u, v, c));
  G[v].push_back(edges.size());
  edges.push_back(Edge(v, u, 0));
}
int was[M], dist[M], s, t;
long long Cut(int u, long long flow) {
  if (u == t) return flow;
  if (!flow || dist[u] >= dist[t]) return 0;
  int ans = 0;
  for (; was[u] < G[u].size(); was[u]++) {
    int e = G[u][was[u]];
    int v = edges[e].v;
    long long c = edges[e].c;
    if (!c || dist[v] != dist[u] + 1) continue;
    int cut = Cut(v, min(flow, c));
    edges[e].c -= cut;
    edges[e ^ 1].c += cut;
    ans += cut;
    flow -= cut;
    if (!flow) return ans;
  }
  return ans;
}
bool BFS() {
  for (int i = s; i <= t; i++) dist[i] = inf, was[i] = 0;
  dist[s] = 0;
  queue<int> q;
  q.push(s);
  while (q.size()) {
    int u = q.front();
    q.pop();
    for (int e : G[u]) {
      int v = edges[e].v;
      long long c = edges[e].c;
      if (c && dist[v] > dist[u] + 1) {
        dist[v] = dist[u] + 1;
        q.push(v);
      }
    }
  }
  return dist[t] != inf;
}
long long MaxFlow() {
  long long Flow = 0;
  while (BFS()) Flow += Cut(s, 1e18);
  return Flow;
}
int id[N][N];
int main() {
  int n, h, m, i, j, c = 0;
  scanf("%i %i %i", &n, &h, &m);
  s = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= h; j++) {
      id[i][j] = ++c;
      if (j > 0) AddEdge(id[i][j - 1], id[i][j], h * h - (j - 1) * (j - 1));
    }
    AddEdge(s, id[i][0], inf);
  }
  t = c + m + 1;
  for (i = 1; i <= m; i++) {
    int l, r, x, p;
    scanf("%i %i %i %i", &l, &r, &x, &p);
    c++;
    AddEdge(c, t, p);
    if (x != h)
      for (j = l; j <= r; j++) AddEdge(id[j][x + 1], c, inf);
  }
  printf("%lld\n", (long long)n * h * h - MaxFlow());
  return 0;
}
