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>
#include <string>
#define ft first
#define sc second
#define pt(sth) cout << sth << "\n"
#define chmax(a, b) (a)=max(a, b)
#define chmin(a, b) (a)=min(a, b)
#define moC(a, s, b) (a)=((a)s(b)+MOD)%MOD
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
static const ll INF=1e18;
static const ll MAX=101010;
static const ll MOD=1e9+7;

/*
 for(i=0; i<N; i++)
   cin >> a[i];
*/

ll N, M, Q;
ll a[55], b[55], c[55], d[55];

ll dfs(vector<ll> v, ll n) {
  ll i;
  
  if(n==N) {
    ll t=0;
    for(i=0; i<Q; i++) if(v[b[i]]-v[a[i]]==c[i]) t+=d[i];
    return t;
  }
  
  ll res=0;
  
  for(i=(n==0) ? 1:v[n-1]; i<=M; i++) {
    vector<ll> u(n);
    copy(v.begin(), v.end(), u.begin());
    u.push_back(i);
    chmax(res, dfs(u, n+1));
  }
  
  return res;
}

int main(void) {
  ll i, j, k;
  
  cin >> N >> M >> Q;
  for(i=0; i<Q; i++) {
    cin >> a[i] >> b[i] >> c[i] >> d[i];
    a[i]--; b[i]--;
  }
  
  vector<ll> v;
  pt(dfs(v, 0));
}