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 = 1000005;
const int M = 1000005;
const int inf = 0x3f3f3f3f;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
struct edge {
  int to, next;
  long long cost;
} g[M * 2];
int head[N], d[N], cnt = 1, now[N], a[N], b[N];
queue<int> q;
void add(int u, int v, int w) {
  g[++cnt].to = v;
  g[cnt].next = head[u];
  head[u] = cnt;
  g[cnt].cost = w;
  g[++cnt].to = u;
  g[cnt].next = head[v];
  head[v] = cnt;
  g[cnt].cost = 0;
}
bool bfs(int s, int t) {
  memset(d, 0, sizeof(d));
  while (q.size()) q.pop();
  q.push(s);
  d[s] = 1;
  now[s] = head[s];
  while (q.size()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = g[i].next) {
      if (g[i].cost && !d[g[i].to]) {
        q.push(g[i].to);
        now[g[i].to] = head[g[i].to];
        d[g[i].to] = d[x] + 1;
        if (g[i].to == t) return 1;
      }
    }
  }
  return 0;
}
int dinic(int x, int t, int flow) {
  if (x == t) return flow;
  long long minn = flow, k;
  for (int i = now[x]; i && minn; i = g[i].next) {
    now[x] = i;
    if (g[i].cost && d[g[i].to] == d[x] + 1) {
      k = dinic(g[i].to, t, min(minn, g[i].cost));
      if (!k) {
        d[g[i].to] = 0;
        continue;
      }
      g[i].cost -= k;
      g[i ^ 1].cost += k;
      minn -= k;
      if (!g[i].cost) break;
    }
  }
  return flow - minn;
}
long long solve(int s, int t) {
  long long ans = 0, flow = 0;
  while (bfs(s, t)) {
    while (flow = dinic(s, t, inf)) ans += flow;
  }
  return ans;
}
int main() {
  int n, h, m;
  scanf("%d%d%d", &n, &h, &m);
  long long ans = n * h * h;
  int s = n * (h + 1) + 1, t = n * (h + 1) + 2;
  int sum = t;
  for (int i = 1; i <= n; i++) {
    add(s, (i - 1) * (h + 1) + 1, inf);
    for (int j = 1; j <= h; j++) {
      add((i - 1) * (h + 1) + j, (i - 1) * (h + 1) + j + 1,
          h * h - (j - 1) * (j - 1));
    }
    add(i * (h + 1), t, 0);
  }
  for (int i = 1; i <= m; i++) {
    int l, r, x, c;
    scanf("%d%d%d%d", &l, &r, &x, &c);
    if (x == h) continue;
    x += 2;
    sum++;
    for (int j = l; j <= r; j++) {
      add((j - 1) * (h + 1) + x, sum, inf);
    }
    add(sum, t, c);
  }
  printf("%lld\n", ans - solve(s, t));
}
