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 = 3575;
const int maxm = maxn * 4 + maxn * 100;
const int INF = 0x3f3f3f3f;
const int inf = 0x3f3f3f3f;
struct Edge {
  int to, next, cap, flow;
} edge[maxm];
int tol, head[maxn];
int h;
void init() {
  tol = 2;
  memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int w, int rw = 0) {
  edge[tol].to = v;
  edge[tol].cap = w;
  edge[tol].flow = 0;
  edge[tol].next = head[u];
  head[u] = tol++;
  edge[tol].to = u;
  edge[tol].cap = rw;
  edge[tol].flow = 0;
  edge[tol].next = head[v];
  head[v] = tol++;
}
int Q[maxn];
int dep[maxn], cur[maxn], sta[maxn];
bool bfs(int s, int t, int n) {
  int front = 0, tail = 0;
  memset(dep, -1, sizeof(dep[0]) * (n + 1));
  dep[s] = 0;
  Q[tail++] = s;
  while (front < tail) {
    int u = Q[front++];
    for (int i = head[u]; i != -1; i = edge[i].next) {
      int v = edge[i].to;
      if (edge[i].cap > edge[i].flow && dep[v] == -1) {
        dep[v] = dep[u] + 1;
        if (v == t) {
          return true;
        }
        Q[tail++] = v;
      }
    }
  }
  return false;
}
int dinic(int s, int t, int n) {
  int maxflow = 0;
  while (bfs(s, t, n)) {
    for (int i = 0; i < n; i++) {
      cur[i] = head[i];
    }
    int u = s, tail = 0;
    while (cur[s] != -1) {
      if (u == t) {
        int tp = INF;
        for (int i = tail - 1; i >= 0; i--) {
          tp = min(tp, edge[sta[i]].cap - edge[sta[i]].flow);
        }
        maxflow += tp;
        for (int i = tail - 1; i >= 0; i--) {
          edge[sta[i]].flow += tp;
          edge[sta[i] ^ 1].flow -= tp;
          if (edge[sta[i]].cap - edge[sta[i]].flow == 0) {
            tail = i;
          }
        }
        u = edge[sta[tail] ^ 1].to;
      } else if (cur[u] != -1 && edge[cur[u]].cap > edge[cur[u]].flow &&
                 dep[u] + 1 == dep[edge[cur[u]].to]) {
        sta[tail++] = cur[u];
        u = edge[cur[u]].to;
      } else {
        while (u != s && cur[u] == -1) {
          u = edge[sta[--tail] ^ 1].to;
        }
        cur[u] = edge[cur[u]].next;
      }
    }
  }
  return maxflow;
}
int Point(int x, int y) { return (x - 1) * (h + 2) + y; }
int main() {
  int n, m;
  cin >> n >> h >> m;
  int T = n * (h + 2) + 2 + m, S = T - 1;
  init();
  for (int i = 1; i <= n; i++) {
    addedge(S, Point(i, 1), inf);
    addedge(Point(i, h + 2), T, inf);
  }
  for (int i = 2; i <= h + 2; i++) {
    for (int j = 1; j <= n; j++) {
      addedge(Point(j, i - 1), Point(j, i), h * h - (i - 2) * (i - 2));
    }
  }
  int cnt = n * (h + 2);
  int l, r, x, c;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d%d", &l, &r, &x, &c);
    ++cnt;
    for (int j = l; j <= r; j++) {
      addedge(Point(j, x + 2), cnt, inf);
    }
    addedge(cnt, T, c);
  }
  long long ans = n * h * h;
  ans -= dinic(S, T, T + 1);
  cout << ans;
  return 0;
}
