Problem:
Score : 300 points

Problem Statement

Given are positive integers N, M, Q, and Q quadruples of integers ( a_i , b_i , c_i , d_i ).

Consider a sequence A satisfying the following conditions:

A is a sequence of N positive integers.

1 \leq A_1 \leq A_2 \le \cdots \leq A_N \leq M.

Let us define a score of this sequence as follows:

The score is the sum of d_i over all indices i such that A_{b_i} - A_{a_i} = c_i. (If there is no such i, the score is 0.)

Find the maximum possible score of A.

Constraints

All values in input are integers.

2 ≤ N ≤ 10

1 \leq M \leq 10

1 \leq Q \leq 50

1 \leq a_i < b_i \leq N ( i = 1, 2, ..., Q )

0 \leq c_i \leq M - 1 ( i = 1, 2, ..., Q )

(a_i, b_i, c_i) \neq (a_j, b_j, c_j) (where i \neq j)

1 \leq d_i \leq 10^5 ( i = 1, 2, ..., Q )

Input

Input is given from Standard Input in the following format:

N M Q
a_1 b_1 c_1 d_1
:
a_Q b_Q c_Q d_Q

Output

Print the maximum possible score of A.

Sample Input 1

3 4 3
1 3 3 100
1 2 2 10
2 3 2 10

Sample Output 1

110

When A = \{1, 3, 4\}, its score is 110. Under these conditions, no sequence has a score greater than 110, so the answer is 110.

Sample Input 2

4 6 10
2 4 1 86568
1 4 0 90629
2 3 0 90310
3 4 1 29211
3 4 3 78537
3 4 2 8580
1 2 1 96263
1 4 2 2156
1 2 0 94325
1 4 3 94328

Sample Output 2

357500

Sample Input 3

10 10 1
1 10 9 1

Sample Output 3

1

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define FOR(var, begin, end) for (int var = (begin); var <= (end); var++)
#define RFOR(var, begin, end) for (int var = (begin); var >= (end); var--)
#define REP(var, length) FOR(var, 0, length - 1)
#define RREP(var, length) RFOR(var, length - 1, 0)
#define EACH(value, var) for (auto value : var)
#define SORT(var) sort(var.begin(), var.end())
#define REVERSE(var) reverse(var.begin(), var.end())
#define RSORT(var) \
  SORT(var);       \
  REVERSE(var)
#define OPTIMIZE_STDIO         \
  ios::sync_with_stdio(false); \
  cin.tie(0);                  \
  cout.precision(10);          \
  cout << fixed
#define endl '\n'

const int INF = 1e9;
const ll MOD = 1e9 + 7;
const ll LINF = 1e18;
struct mint {
  ll x;
  mint() : x(0) {
  }
  mint(ll x) : x((x % MOD + MOD) % MOD) {
  }
  mint& operator+=(const mint& r) {
    if ((x += r.x) >= MOD) {
      x -= MOD;
    }
    return *this;
  }
  mint& operator-=(const mint& r) {
    if ((x -= r.x) < 0) {
      x += MOD;
    }
    return *this;
  }
  mint& operator*=(const mint& r) {
    if ((x *= r.x) > MOD) {
      x %= MOD;
    }
    return *this;
  }
  mint& operator++() {
    if ((++x) >= MOD) {
      x -= MOD;
    }
    return *this;
  }
  mint operator++(int) {
    mint ret = x;
    if ((++x) >= MOD) {
      x -= MOD;
    }
    return ret;
  }
  mint& operator--() {
    if ((--x) < 0) {
      x += MOD;
    }
    return *this;
  }
  mint operator--(int) {
    mint ret = x;
    if ((--x) < 0) {
      x += MOD;
    }
    return ret;
  }
  mint operator+(const mint& r) {
    mint ret;
    return ret = x + r.x;
  }
  mint operator-(const mint& r) {
    mint ret;
    return ret = x - r.x;
  }
  mint operator*(const mint& r) {
    mint ret;
    return ret = x * r.x;
  }
  mint operator-() {
    return mint() - *this;
  }
  bool operator<(const mint& a) {
    return x < a.x;
  }
  bool operator>(const mint& a) {
    return x > a.x;
  }
  bool operator<=(const mint& a) {
    return x <= a.x;
  }
  bool operator>=(const mint& a) {
    return x >= a.x;
  }
  bool operator==(const mint& a) {
    return x == a.x;
  }
  bool operator!=(const mint& a) {
    return x != a.x;
  }
};
ostream& operator<<(ostream& os, const mint& r) {
  os << r.x;
  return os;
}
ostream& operator>>(ostream& os, const mint& r) {
  os >> r.x;
  return os;
}

int n, m, q;
vector<int> a, b, c, d;
int ans;

void dfs(vector<int> buf) {
  if (buf.size() == n + 1) {
    int now = 0;
    REP(i, q) {
      if ((buf[b[i]] - buf[a[i]]) == c[i]) now += d[i];
    }
    ans = max(ans, now);
    return;
  }

  buf.push_back(buf.back());
  while (buf.back() <= m) {
    dfs(buf);
    buf.back()++;
  }
}

void solve(istream& cin, ostream& cout) {
  cin >> n >> m >> q;
  a = b = c = d = vector<int>(q);
  REP(i, q) {
    cin >> a[i] >> b[i] >> c[i] >> d[i];
  }
  dfs(vector<int>(1, 1));
  cout << ans << endl;
}

#ifndef TEST
int main() {
  OPTIMIZE_STDIO;
  solve(cin, cout);
  return 0;
}
#endif