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 inf = 1e9, N = 3e3, M = 1e6;
int a[55][55], ver[M], head[N], cnt = 1, edge[M], tot, nxt[M], d[N], S, T, n, m,
                                h;
queue<int> q;
void Add_edge(int x, int y, int z) {
  ver[++cnt] = y;
  nxt[cnt] = head[x];
  head[x] = cnt;
  edge[cnt] = z;
  ver[++cnt] = x;
  nxt[cnt] = head[y];
  head[y] = cnt;
  edge[cnt] = 0;
}
bool Bfs() {
  memset(d, 0, sizeof(d));
  d[S] = 1;
  while (q.size()) q.pop();
  q.push(S);
  while (q.size()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; i; i = nxt[i])
      if (!d[ver[i]] && edge[i]) {
        d[ver[i]] = d[u] + 1;
        q.push(ver[i]);
        if (ver[i] == T) return 1;
      }
  }
  return 0;
}
int Dfs(int u, int flow) {
  if (u == T) return flow;
  int rest = flow, w;
  for (int i = head[u]; i; i = nxt[i])
    if (edge[i] && d[ver[i]] == d[u] + 1) {
      w = Dfs(ver[i], min(rest, edge[i]));
      if (!w) d[ver[i]] = 0;
      edge[i] -= w;
      edge[i ^ 1] += w;
      rest -= w;
    }
  return flow - rest;
}
int Dinic() {
  int mxflow = 0, flow;
  while (Bfs())
    while (flow = Dfs(S, inf)) mxflow += flow;
  return mxflow;
}
int main() {
  scanf("%d%d%d", &n, &h, &m);
  S = 0;
  T = n * (h + 1) + m + 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= h; ++j) a[i][j] = ++tot;
  for (int i = 1; i <= n; ++i) Add_edge(S, a[i][0], inf);
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < h; j++) Add_edge(a[i][j], a[i][j + 1], h * h - j * j);
  for (int l, r, x, c; m--;) {
    scanf("%d%d%d%d", &l, &r, &x, &c);
    if (x >= h) continue;
    ++x;
    ++tot;
    for (int i = l; i <= r; i++) Add_edge(a[i][x], tot, inf);
    Add_edge(tot, T, c);
  }
  printf("%d\n", n * h * h - Dinic());
}
