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;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
template <class Iter>
void rprintf(const char *fmt, Iter begin, Iter end) {
  for (bool sp = 0; begin != end; ++begin) {
    if (sp)
      putchar(' ');
    else
      sp = true;
    printf(fmt, *begin);
  }
  putchar('\n');
}
int N, H, M;
int L[55], R[55], X[55], C[55];
long long dp[55][55][55];
long long D[55];
void MAIN() {
  scanf("%d%d%d", &N, &H, &M);
  for (int i = 0, i_len = (M); i < i_len; ++i) {
    scanf("%d%d%d%d", L + i, R + i, X + i, C + i), L[i]--;
  }
  for (int l = N; l--;) {
    for (int h = 0, h_len = (H + 1); h < h_len; ++h) {
      memset(D, 0, sizeof D);
      for (int i = 0, i_len = (M); i < i_len; ++i)
        if (l == L[i] && l + 1 == R[i]) {
          D[X[i] + 1] += C[i];
        }
      for (int a = 0, a_len = (h + 1); a < a_len; ++a) {
        D[a + 1] += D[a];
        long long tmp = a * a - D[a];
        amax(dp[l][l + 1][h], tmp);
      }
    }
    for (int k = l; k < N; k++)
      for (int r = k + 1; r <= N; r++)
        for (int h = 0, h_len = (H + 1); h < h_len; ++h) {
          memset(D, 0, sizeof D);
          for (int i = 0, i_len = (M); i < i_len; ++i)
            if (l <= L[i] && L[i] <= k && k < R[i] && R[i] <= r) {
              D[X[i] + 1] += C[i];
            }
          for (int a = 0, a_len = (h + 1); a < a_len; ++a) {
            D[a + 1] += D[a];
            long long tmp = a * a - D[a] + dp[l][k][a] + dp[k + 1][r][a];
            amax(dp[l][r][h], tmp);
          }
        }
  }
  long long ans = dp[0][N][H];
  printf("%lld\n", ans);
}
int main() {
  int TC = 1;
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}
