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 = 52;
const int Inf = 1000000000;
int n, h, m;
int L[Maxn], R[Maxn], X[Maxn], C[Maxn];
int seq[Maxn];
int gdp[Maxn][Maxn], dp[Maxn][Maxn];
int res;
bool Less(const int &a, const int &b) { return X[a] < X[b]; }
void Add(int h) {
  for (int l = 0; l < n; l++) {
    for (int r = l; r < n; r++) dp[l][r] = gdp[l][r];
    dp[l][l] = max(dp[l][l], h * h);
    for (int r = l; r < n; r++) {
      if (r + 1 < n) dp[l][r + 1] = max(dp[l][r + 1], dp[l][r] + h * h);
      for (int r2 = r + 1; r2 < n; r2++)
        dp[l][r2] = max(dp[l][r2], dp[l][r] + gdp[r + 1][r2]);
    }
  }
}
int main() {
  scanf("%d %d %d", &n, &h, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d %d", &L[i], &R[i], &X[i], &C[i]);
    L[i]--;
    R[i]--;
    seq[i] = i;
    res -= C[i];
  }
  sort(seq, seq + m, Less);
  fill((int *)gdp, (int *)gdp + Maxn * Maxn, -Inf);
  for (int i = 0; i < m; i++) {
    int ind = seq[i];
    Add(X[ind]);
    for (int l = 0; l <= L[ind]; l++)
      for (int r = R[ind]; r < n; r++)
        gdp[l][r] = max(gdp[l][r], dp[l][r] + C[ind]);
  }
  Add(h);
  printf("%d\n", res + dp[0][n - 1]);
  return 0;
}
