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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 200010;
const int INF = 2147483600;
int N, H, M;
int S = 0, T;
int dis[MAXN + 1];
int Node[MAXN << 1], Next[MAXN << 1], Root[MAXN + 1], cnt, C[MAXN << 1];
inline void addedge(int u, int v, int w) {
  Node[cnt] = v;
  Next[cnt] = Root[u];
  Root[u] = cnt;
  C[cnt] = w;
  ++cnt;
}
inline void insert(int u, int v, int w) {
  addedge(u, v, w);
  addedge(v, u, 0);
}
inline int Id(int x, int y) { return (x - 1) * (H + 1) + y + 1; }
inline bool BFS() {
  queue<int> que;
  que.push(S);
  for (int i = 0; i <= T; i++) dis[i] = -1;
  dis[S] = 0;
  while (!que.empty()) {
    int k = que.front();
    que.pop();
    for (int x = Root[k]; x != -1; x = Next[x]) {
      int v = Node[x];
      if (C[x] > 0 && dis[v] == -1) dis[v] = dis[k] + 1, que.push(v);
    }
  }
  return (dis[T] != -1);
}
inline int DFS(int k, int t) {
  if (k == T) return t;
  int sum = 0;
  for (int x = Root[k]; x != -1; x = Next[x]) {
    int v = Node[x];
    if (C[x] > 0 && dis[v] == dis[k] + 1) {
      int tmp = DFS(v, min(t, C[x]));
      if (!tmp) continue;
      C[x] -= tmp;
      C[x ^ 1] += tmp;
      sum += tmp;
      t -= tmp;
      if (!t) return sum;
    }
  }
  if (!sum) dis[k] = -1;
  return sum;
}
int main() {
  memset(Root, -1, sizeof(Root));
  N = read(), H = read(), M = read();
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j < H; j++) insert(Id(i, j + 1) + M, Id(i, j) + M, INF);
    for (int j = 1; j <= H; j++) insert(S, Id(i, j) + M, 2 * (j - 1) + 1);
  }
  int ans = N * H * H;
  T = Id(N, H) + M + 1;
  for (int i = 1; i <= M; i++) {
    int l = read(), r = read(), x = read(), c = read();
    if (x == H) continue;
    insert(i, T, c);
    for (int j = l; j <= r; j++) insert(Id(j, x + 1) + M, i, INF);
  }
  while (BFS()) ans -= DFS(S, INF);
  printf("%d\n", ans);
}
