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>
const char nl = '\n';
using namespace std;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int N = 55;
struct restrict {
  int l, r, x, c;
};
int n, h, m;
vector<restrict> rests;
int dp[N][N][N];
int solve(int l, int r, int x) {
  if (l > r) return 0;
  if (dp[l][r][x] != -1) return dp[l][r][x];
  int ans = 0;
  for (int k = 1; k < x; k++) {
    ans = max(ans, solve(l, r, k));
  }
  for (int i = l; i <= r; i++) {
    int pen = 0;
    for (const auto& rs : rests) {
      if (rs.l > i || rs.r < i) continue;
      if (rs.l < l || rs.r > r) continue;
      if (rs.x >= x) continue;
      pen += rs.c;
    }
    ans = max(ans, solve(l, i - 1, x) + solve(i + 1, r, x) + x * x - pen);
  }
  return dp[l][r][x] = ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> h >> m;
  rests.resize(m);
  for (int i = 0; i < m; i++) {
    cin >> rests[i].l >> rests[i].r >> rests[i].x >> rests[i].c;
  }
  memset(dp, -1, sizeof dp);
  cout << solve(1, n, h) << nl;
  return 0;
}
