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 MAXN = (long long)53;
const int infint = (long long)1e9 + 3;
const int MOD = (long long)1e9 + 7;
const long long inf = (long long)1e18 + 3;
int n, x, m, dp[MAXN][MAXN][MAXN], l[MAXN], r[MAXN], h[MAXN], c[MAXN];
vector<int> query[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> x >> m;
  for (int i = 0; i < m; i++) {
    cin >> l[i] >> r[i] >> h[i] >> c[i];
    query[l[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++)
    for (int mx = 0; mx <= x; mx++) {
      if (mx) dp[i][i][mx] = dp[i][i][mx - 1];
      int ans = mx * mx;
      for (auto idx : query[i])
        if (r[idx] == i && h[idx] < mx) ans -= c[idx];
      dp[i][i][mx] = max(dp[i][i][mx], ans);
    }
  for (int len = 1; len < n; len++)
    for (int i = 1; i + len <= n; i++)
      for (int mx = 1; mx <= x; mx++) {
        int j = i + len;
        if (mx) dp[i][j][mx] = dp[i][j][mx - 1];
        for (int k = i; k <= j; k++) {
          int ans = mx * mx;
          for (int lft = i; lft <= k; lft++)
            for (auto idx : query[lft])
              if (r[idx] >= k && r[idx] <= j && h[idx] < mx) ans -= c[idx];
          if (k != i) ans += dp[i][k - 1][mx - 1];
          if (k != j) ans += dp[k + 1][j][mx];
          dp[i][j][mx] = max(dp[i][j][mx], ans);
        }
      }
  cout << dp[1][n][x];
}
