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 MAXV = 1e4 + 10;
const int MAXE = 1e5 + 5;
const long long INF = 0x3f3f3f3f3f3f3f3f;
long long tot;
long long head[MAXV], level[MAXV], iter[MAXV];
struct Edge {
  long long v, cap, nxt;
  Edge() {}
  Edge(long long v, long long cap, long long nxt) : v(v), cap(cap), nxt(nxt) {}
} edge[MAXE << 1];
void init() {
  tot = 0;
  memset(head, -1, sizeof(head));
}
void add_edge(long long u, long long v, long long c) {
  edge[tot] = Edge(v, c, head[u]);
  head[u] = tot++;
  edge[tot] = Edge(u, 0, head[v]);
  head[v] = tot++;
}
void bfs(long long s) {
  memset(level, -1, sizeof(level));
  level[s] = 0;
  queue<long long> que;
  que.push(s);
  while (!que.empty()) {
    long long u = que.front();
    que.pop();
    for (long long i = head[u]; ~i; i = edge[i].nxt) {
      long long v = edge[i].v;
      long long c = edge[i].cap;
      if (c > 0 && level[v] < 0) {
        level[v] = level[u] + 1;
        que.push(v);
      }
    }
  }
}
long long dfs(long long u, long long t, long long f) {
  if (u == t) return f;
  for (long long &i = iter[u]; ~i; i = edge[i].nxt) {
    long long v = edge[i].v;
    long long c = edge[i].cap;
    if (c > 0 && level[u] < level[v]) {
      long long d = dfs(v, t, min(f, c));
      if (d > 0) {
        edge[i].cap -= d;
        edge[i ^ 1].cap += d;
        return d;
      }
    }
  }
  return 0;
}
long long max_flow(long long s, long long t) {
  long long flow = 0;
  while (1) {
    bfs(s);
    if (level[t] < 0) return flow;
    long long f;
    memcpy(iter, head, sizeof(head));
    while (f = dfs(s, t, INF)) flow += f;
  }
}
int node[55][55];
int main() {
  init();
  int n, h, m;
  scanf("%d%d%d", &n, &h, &m);
  long long ans = 1LL * n * h * h;
  int tot = 0;
  int s = ++tot;
  int t = ++tot;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= h; j++) {
      node[i][j] = ++tot;
      if (j != 1) {
        add_edge(node[i][j], node[i][j - 1], INF);
      }
      add_edge(s, node[i][j], 2 * j - 1);
    }
  }
  for (int i = 1; i <= m; i++) {
    ++tot;
    int l, r, x, c;
    scanf("%d%d%d%d", &l, &r, &x, &c);
    if (x >= h) continue;
    for (int j = l; j <= r; j++) {
      add_edge(node[j][x + 1], tot, INF);
    }
    add_edge(tot, t, c);
  }
  printf("%lld", ans - max_flow(s, t));
  return 0;
}
