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;
void Freopen() {
  freopen(
      "CF1146G Zoning Restrictions"
      ".in",
      "r", stdin);
  freopen(
      "CF1146G Zoning Restrictions"
      ".out",
      "w", stdout);
}
int read() {
  int g = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    g = g * 10 + ch - '0';
    ch = getchar();
  }
  return g * f;
}
const int N = 55;
const int INF = 1e9;
struct node {
  int to, ne, c;
} e[(N * N) << 2];
int S, T, dis[N * N], h[N * N], cur[N * N], cnt = 1;
int n, hig, m;
void add(int x, int y, int c) {
  e[++cnt] = node{y, h[x], c};
  h[x] = cnt;
  e[++cnt] = node{x, h[y], 0};
  h[y] = cnt;
}
bool bfs(int s, int t) {
  for (int i = (1); i <= (T); i++) dis[i] = 0;
  dis[s] = 1;
  queue<int> q;
  q.push(s);
  while (q.size()) {
    int x = q.front();
    q.pop();
    if (x == t) return true;
    for (int i = h[x]; i; i = e[i].ne) {
      int y = e[i].to;
      if (!e[i].c) continue;
      if (dis[y]) continue;
      dis[y] = dis[x] + 1;
      q.push(y);
    }
  }
  return false;
}
int dfs(int x, int maxn, int t) {
  if (x == t || maxn == 0) return maxn;
  int ret = 0, dlt = 0;
  for (int &i = cur[x]; i; i = e[i].ne) {
    int y = e[i].to;
    if (dis[y] != dis[x] + 1) continue;
    if (!e[i].c) continue;
    dlt = dfs(y, min(e[i].c, maxn), t);
    if (!dlt) dis[y] = 0;
    e[i].c -= dlt, e[i ^ 1].c += dlt;
    maxn -= dlt, ret += dlt;
    if (!maxn) return ret;
  }
  return ret;
}
int id(int dep, int x) { return (dep - 1) * n + x; }
signed main() {
  n = read(), hig = read(), m = read();
  S = (hig + 1) * n + m + 1, T = S + 1;
  for (int i = (1); i <= (n); i++) {
    add(S, id(1, i), hig * hig);
    for (int op = (2); op <= (hig + 1); op++)
      add(id(op - 1, i), id(op, i), hig * hig - (op - 1) * (op - 1));
  }
  for (int i = (1); i <= (m); i++) {
    int l = read(), r = read(), x = read(), c = read();
    for (int j = (l); j <= (r); j++) add(id(x + 1, j), (hig + 1) * n + i, INF);
    add((hig + 1) * n + i, T, c);
  }
  int ans = 0;
  while (bfs(S, T)) {
    for (int i = (1); i <= (T); i++) cur[i] = h[i];
    ans += dfs(S, INF, T);
  }
  cout << 1ll * n * hig * hig - ans;
  return signed();
}
