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;
int head[222222], to[222222], cap[222222], dist[222222], nxt[222222],
    cur[222222];
int n, m, s, t, cnt, x, y, z, h, ans, l, r, c;
void addedge(int s, int t, int cp) {
  cnt++;
  nxt[cnt] = head[s];
  head[s] = cnt;
  to[cnt] = t;
  cap[cnt] = cp;
}
void bfs() {
  memset(dist, -1, sizeof(dist));
  queue<int> q;
  q.push(s);
  dist[s] = 0;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = nxt[i]) {
      if (!cap[i]) continue;
      if (!~dist[to[i]]) {
        dist[to[i]] = dist[x] + 1;
        q.push(to[i]);
      }
    }
  }
}
int dfs(int i, int f) {
  if (i == t) return f;
  int tmp = f;
  for (int &j = cur[i]; j; j = nxt[j]) {
    if (!cap[j] || dist[to[j]] != dist[i] + 1) continue;
    int d = dfs(to[j], min(cap[j], tmp));
    if (d) {
      cap[j] -= d;
      cap[j ^ 1] += d;
      tmp -= d;
      if (!tmp) break;
    } else
      dist[to[j]] = 0;
  }
  return f - tmp;
}
int dinic() {
  int res = 0;
  while (1) {
    for (int i = 1; i <= t; i++) cur[i] = head[i];
    bfs();
    if (!~dist[t]) break;
    res += dfs(s, 1e9);
  }
  return res;
}
int get(int x, int y) { return (x - 1) * h + y; }
int main() {
  scanf("%d%d%d", &n, &h, &m);
  s = n * h + m + 1;
  t = s + 1;
  cnt = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= h; j++) {
      ans += (j * 2 - 1);
      addedge(s, get(i, j), j * 2 - 1);
      addedge(get(i, j), s, 0);
    }
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d%d", &l, &r, &x, &c);
    addedge(n * h + i, t, c);
    addedge(t, n * h + i, 0);
    for (int j = l; j <= r; j++) {
      for (int k = x + 1; k <= h; k++) {
        addedge(get(j, k), n * h + i, inf);
        addedge(n * h + i, get(j, k), 0);
      }
    }
  }
  printf("%d\n", ans - dinic());
  return 0;
}
