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 = 3000, M = 1000000, inf = 0x3f3f3f3f;
struct node {
  int to, nxt, flow;
} edge[M];
int head[N], cur[N], dep[N], a[N][N], cnt, n, S, T, h, m, tot, ans;
void pre() {
  memset(head, -1, sizeof(head));
  cnt = -1;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + ch - '0';
  return x * f;
}
void add(int u, int v, int w) {
  edge[++cnt] = (node){v, head[u], w};
  head[u] = cnt;
  edge[++cnt] = (node){u, head[v], 0};
  head[v] = cnt;
}
bool bfs() {
  memset(dep, 0, sizeof(dep));
  dep[S] = 1;
  queue<int> q;
  q.push(S);
  memcpy(cur, head, sizeof(head));
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; ~i; i = edge[i].nxt) {
      int v = edge[i].to, f = edge[i].flow;
      if (f && !dep[v]) dep[v] = dep[u] + 1, q.push(v);
    }
  }
  return dep[T];
}
int dfs(int u, int lim) {
  if (!lim || u == T) return lim;
  int addflow = 0, flow;
  for (int i = cur[u]; ~i; i = edge[i].nxt) {
    cur[u] = i;
    int v = edge[i].to, f = edge[i].flow;
    if (dep[v] == dep[u] + 1 && (flow = dfs(v, min(lim, f)))) {
      addflow += flow;
      lim -= flow;
      edge[i].flow -= flow;
      edge[i ^ 1].flow += flow;
      if (!lim) break;
    }
  }
  return addflow;
}
int Dinic() {
  int maxflow = 0;
  while (bfs()) maxflow += dfs(S, inf);
  return maxflow;
}
int main() {
  pre();
  n = read();
  h = read();
  m = read();
  tot = 0;
  ans = n * h * h;
  S = n * (h + 1) + m + 1;
  T = S + 1;
  for (register int i = (1); i <= (n); ++i)
    for (register int j = (0); j <= (h); ++j) a[i][j] = ++tot;
  for (register int i = (1); i <= (n); ++i) add(S, a[i][0], inf);
  for (register int i = (1); i <= (n); ++i)
    for (register int j = (1); j <= (h); ++j)
      add(a[i][j - 1], a[i][j], h * h - (j - 1) * (j - 1));
  while (m--) {
    int l = read(), r = read(), x = read(), c = read();
    if (x == h) continue;
    tot++;
    x++;
    for (register int i = (l); i <= (r); ++i) add(a[i][x], tot, inf);
    add(tot, T, c);
  }
  printf("%d\n", ans - Dinic());
}
