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)

/* ****revpro**** */
// #pragma GCC target ("avx2")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")
#include<bits/stdc++.h>
#define ll          long long
#define pb          push_back
#define endl     '\n'
#define pii         pair<ll,ll>
#define vi          vector<ll>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define rep(i,a,b)  for(ll i=a;i<b;i++)
#define repr(i,a,b) for(ll i=b-1;i>=a;i--)    
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define mp          make_pair
#define present(c,x) ((c).find(x) != (c).end())
#define cpresent(c,x) (find(all(c),x) != (c).end())
#define mii      map<ll,ll>
using namespace std;

#define TIME        cerr << "\nTime elapsed: " << setprecision(5) <<1000.0 * clock() / CLOCKS_PER_SEC << "ms\n";
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define N  100005
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;  
#define ordered_set tree<double, null_type,less<double>, rb_tree_tag,tree_order_statistics_node_update> 
ll dp[11][11][11];
map <ll,ll> ans,vis;
void solve()
{
  ll n,m,tt;
  cin>>n>>m>>tt;
  rep(i,0,tt)
  {
    ll a,b,c,d;
    cin>>a>>b>>c>>d;
    dp[a-1][b-1][c]+=d;
  }
  ans[0]=0;
  priority_queue <ll> q;
  q.push(0);
  vi v;
  ll x,x1,st,sum=0;
  ll tmp=0;
  vis[0]=1;
  while(!q.empty())
  {
    v.clear();
    x=q.top();
    x=-x;
    q.pop();
    x1=x;
    while(x1)
    {
      v.pb(x1%11);
      x1/=11;
    }
    if(sz(v)==n) break;
    reverse(all(v));
    st=1;
    if(sz(v)) st=v.back();
    rep(i,st,m+1)
    {
      sum=ans[x];
      rep(j,0,sz(v))
      {
        sum+=dp[j][sz(v)][i-v[j]];
      }
      ans[x*11+i]=max(ans[x*11+i],sum);
      tmp=max(tmp,sum);
      if(vis[-(x*11+i)]==0)
      {
        q.push(-(x*11+i));
        vis[-(x*11+i)]=1;
      }
    }
  }
  cout<<tmp;
}
signed main()
{
  
    /***********************************/
    unsigned seed = std::chrono::system_clock::now().time_since_epoch().count();
    mt19937_64 generator (seed);

  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS=1;
//  cin>>TESTS;
  while(TESTS--)
  {
    solve();
    
  }
  TIME
  return 0;
}