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 = 60;
int n, h, m;
long long dp[maxn][maxn][maxn], cost[maxn][maxn][maxn][maxn];
int l[maxn], r[maxn], c[maxn], x[maxn];
long long calc(int lf, int rg, int hg) {
  if (rg < lf) return 0;
  long long ans = 0;
  if (dp[lf][rg][hg] != -1) {
    return dp[lf][rg][hg];
  }
  for (int len = 0; len <= hg; len++) {
    for (int mid = lf; mid <= rg; mid++) {
      ans = max(ans, calc(lf, mid - 1, len) + calc(mid + 1, rg, len) +
                         len * len - cost[lf][rg][mid][len]);
    }
  }
  dp[lf][rg][hg] = ans;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> h >> m;
  for (int i = 1; i <= m; i++) {
    cin >> l[i] >> r[i] >> x[i] >> c[i];
  }
  memset(dp, -1, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      for (int k = i; k <= j; k++)
        for (int len = 0; len <= h; len++) {
          for (int q = 1; q <= m; q++) {
            if (i <= l[q] && l[q] <= k && k <= r[q] && r[q] <= j &&
                len > x[q]) {
              cost[i][j][k][len] += c[q];
            }
          }
        }
    }
  }
  long long ans = calc(1, n, h);
  cout << ans;
  return 0;
}
