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 = 2800;
const int maxm = 30005;
const int inf = 1 << 30;
int nex[maxm], to[maxm], cap[maxm], from[maxm];
int n, m, h, cnt, head[maxn], d[maxn], sp, tp;
void add(int u, int v, int c) {
  from[cnt] = u, to[cnt] = v, cap[cnt] = c, nex[cnt] = head[u], head[u] = cnt++;
  from[cnt] = v, to[cnt] = u, cap[cnt] = 0, nex[cnt] = head[v], head[v] = cnt++;
}
int bfs() {
  queue<int> q;
  memset(d, -1, sizeof(d));
  d[sp] = 0;
  q.push(sp);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    for (int i = head[cur]; i != -1; i = nex[i]) {
      int u = to[i];
      if (d[u] == -1 && cap[i] > 0) {
        d[u] = d[cur] + 1;
        q.push(u);
      }
    }
  }
  return d[tp] != -1;
}
int dfs(int a, int b) {
  int r = 0;
  if (a == tp) return b;
  for (int i = head[a]; i != -1 && r < b; i = nex[i]) {
    int u = to[i];
    if (cap[i] > 0 && d[u] == d[a] + 1) {
      int x = min(cap[i], b - r);
      x = dfs(u, x);
      r += x;
      cap[i] -= x;
      cap[i ^ 1] += x;
    }
  }
  if (!r) d[a] = -2;
  return r;
}
int dinic(int sp, int tp) {
  int total = 0, t;
  while (bfs()) {
    while (t = dfs(sp, inf)) total += t;
  }
  return total;
}
int gain(int id, int hei) { return (id - 1) * (h + 1) + hei + 1; }
int sub(int x) { return h * h - x * x; }
int main() {
  cnt = 0;
  memset(head, -1, sizeof(head));
  scanf("%d%d%d", &n, &h, &m);
  int ans = n * h * h, np = (h + 1) * n;
  sp = 0, tp = np + m + 1;
  for (int i = (int)1; i <= (int)n; i++) {
    add(sp, gain(i, 0), inf);
    for (int j = (int)1; j <= (int)h; j++) {
      add(gain(i, j - 1), gain(i, j), sub(j - 1));
    }
  }
  for (int i = (int)1; i <= (int)m; i++) {
    int l, r, x, c;
    scanf("%d%d%d%d", &l, &r, &x, &c);
    if (x == h) continue;
    for (int j = (int)l; j <= (int)r; j++) {
      add(gain(j, x + 1), np + i, inf);
    }
    add(np + i, tp, c);
  }
  printf("%d\n", ans - dinic(sp, tp));
  return 0;
}
