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 = 3005;
const int inf = 1000 * 1000 * 1000;
const int mod = 1000 * 1000 + 3;
int n, h, m;
int S, T;
struct edge {
  int a, b, cap, flow;
};
vector<int> g[N];
vector<edge> G;
int gag[55][55];
void add_edge(int a, int b, int c) {
  edge e1 = {a, b, c, 0};
  edge e2 = {b, a, 0, 0};
  g[a].push_back((int)G.size());
  G.push_back(e1);
  g[b].push_back((int)G.size());
  G.push_back(e2);
}
int d[N];
bool bfs() {
  queue<int> q;
  for (int i = S; i <= T; i++) {
    d[i] = -1;
  }
  q.push(S);
  d[S] = 0;
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < (int)g[v].size(); i++) {
      int id = g[v][i];
      int to = G[id].b;
      if (G[id].cap == G[id].flow) continue;
      if (d[to] == -1) {
        d[to] = d[v] + 1;
        q.push(to);
      }
    }
  }
  return (d[T] != -1);
}
int pos[N];
int dfs(int v, int flow) {
  if (v == T || flow == 0) return flow;
  for (; pos[v] < (int)g[v].size(); pos[v]++) {
    int id = g[v][pos[v]];
    int to = G[id].b;
    if (d[to] != d[v] + 1) continue;
    int x = dfs(to, min(flow, G[id].cap - G[id].flow));
    if (x) {
      G[id].flow += x;
      G[id ^ 1].flow -= x;
      return x;
    }
  }
  return 0;
}
int flow() {
  int answ = 0;
  while (bfs()) {
    for (int i = S; i <= T; i++) {
      pos[i] = 0;
    }
    while (true) {
      int x = dfs(S, 2 * inf);
      if (x == 0) break;
      answ += x;
    }
  }
  return answ;
}
int main() {
  mt19937 myrand(chrono::steady_clock::now().time_since_epoch().count());
  scanf("%d%d%d", &n, &h, &m);
  S = 0;
  T = n * (h + 1) + m + 1;
  int it = 0;
  for (int i = 1; i <= n; i++) {
    ++it;
    add_edge(S, it, inf);
    for (int j = 0; j < h; j++) {
      gag[i][j] = it;
      add_edge(it, it + 1, h * h - j * j);
      ++it;
    }
    gag[i][h] = it;
  }
  for (int i = 1; i <= m; i++) {
    int l, r, x, c;
    scanf("%d%d%d%d", &l, &r, &x, &c);
    ++it;
    add_edge(it, T, c);
    for (int j = l; j <= r; j++) {
      for (int l = x + 1; l <= h; l++) {
        add_edge(gag[j][l], it, inf);
      }
    }
  }
  cout << h * h * n - flow() << endl;
  return 0;
}
