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 inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 55;
int dp[maxn][maxn][maxn];
int l[maxn], r[maxn], x[maxn], c[maxn];
int N, H, M;
int DP(int a, int b, int w) {
  if (a > b || !w) return dp[a][b][w] = 0;
  if (dp[a][b][w] >= 0) return dp[a][b][w];
  int &ans = dp[a][b][w];
  ans = DP(a, b, w - 1);
  int sum[maxn];
  memset((sum), 0, sizeof(sum));
  for (int i = 1; i <= M; i++) {
    if (l[i] >= a && r[i] <= b && w > x[i]) {
      sum[l[i]] += c[i];
      sum[r[i] + 1] -= c[i];
    }
  }
  for (int i = a + 1; i <= b; i++) sum[i] += sum[i - 1];
  for (int i = a; i <= b; i++) {
    ans = max(ans, DP(a, i - 1, w) + DP(i + 1, b, w) + w * w - sum[i]);
  }
  return ans;
}
int main() {
  memset((dp), -1, sizeof(dp));
  scanf("%d%d%d", &N, &H, &M);
  for (int i = 1; i <= M; i++) scanf("%d%d%d%d", &l[i], &r[i], &x[i], &c[i]);
  printf("%d\n", DP(1, N, H));
}
