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 N = 55;
int n, h, m;
struct node {
  int l, r, x, c;
} p[N];
int f[N][N][N], g[N][N][N];
bool vs[N][N];
void dp(int l, int r) {
  vs[l][r] = 1;
  for (int i = l; i <= r; i++) {
    if (i < r && !vs[l][i]) dp(l, i);
    if (i > l && !vs[i][r]) dp(i, r);
  }
  for (int x = 0; x <= h; x++) {
    int s[N] = {0};
    for (int i = 1; i <= m; i++)
      if (p[i].l >= l && p[i].r <= r && p[i].x < x) {
        s[p[i].l] += p[i].c;
        s[p[i].r + 1] -= p[i].c;
      }
    for (int i = l; i <= r; i++) {
      s[i] += s[i - 1];
      f[l][r][x] =
          max(f[l][r][x], g[l][i - 1][x] + g[i + 1][r][x] + x * x - s[i]);
    }
    if (x)
      g[l][r][x] = max(g[l][r][x - 1], f[l][r][x]);
    else
      g[l][r][x] = f[l][r][x];
  }
}
int main() {
  cin >> n >> h >> m;
  for (int i = 1; i <= m; i++)
    scanf("%d %d %d %d", &p[i].l, &p[i].r, &p[i].x, &p[i].c);
  dp(1, n);
  cout << g[1][n][h] << endl;
  return 0;
}
