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;
int n, h, m, sum, f = 1, sok = 1e6;
const int c = 20002;
int kezd, veg, pos[c], szint[c];
vector<int> sz[c], ki[c];
vector<int> s[c];
void add(int a, int b, int c) {
  ki[a].push_back(sz[b].size()), ki[b].push_back(sz[a].size());
  sz[a].push_back(b), sz[b].push_back(a);
  s[a].push_back(c), s[b].push_back(0);
}
int dfs(int a, int ert) {
  if (a == veg) {
    return ert;
  }
  for (; pos[a] < sz[a].size(); pos[a]++) {
    int x = sz[a][pos[a]];
    int suly = s[a][pos[a]];
    if (szint[x] == szint[a] + 1 && suly) {
      int add = dfs(x, min(ert, suly));
      if (add) {
        s[a][pos[a]] -= add;
        s[x][ki[a][pos[a]]] += add;
        return add;
      }
    }
  }
  return 0;
}
bool bfs() {
  for (int i = kezd; i <= veg; i++) {
    szint[i] = 0;
  }
  szint[kezd] = 1;
  queue<int> q;
  q.push(kezd);
  while (q.size() > 0) {
    int a = q.front();
    q.pop();
    for (int i = 0; i < sz[a].size(); i++) {
      int x = sz[a][i];
      int suly = s[a][i];
      if (suly && !szint[x]) {
        szint[x] = szint[a] + 1;
        q.push(x);
      }
    }
  }
  return szint[veg];
}
int dinic() {
  int ans = 0;
  while (bfs()) {
    for (int i = kezd; i <= veg; i++) {
      pos[i] = 0;
    }
    while (int add = dfs(kezd, sok)) {
      ans += add;
    }
  }
  return ans;
}
int main() {
  cin >> n >> h >> m;
  sum = n * h * h, veg = n * (h + 1) + m + 1;
  for (int i = 1; i <= n; i++) {
    add(kezd, f, h * h);
    f++;
    for (int j = 0; j < h; j++) {
      add(f - 1, f, h * h - j * j);
      f++;
    }
  }
  for (int i = 1; i <= m; i++) {
    int l, r, kr, ert;
    cin >> l >> r >> kr >> ert;
    add(f, veg, ert);
    if (kr < h) {
      for (int j = l; j <= r; j++) {
        add((j - 1) * (h + 1) + kr + 2, f, sok);
      }
    }
    f++;
  }
  cout << sum - dinic();
  return 0;
}
