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 <ext/pb_ds/assoc_container.hpp> 
using namespace __gnu_pbds;
using namespace std;

typedef long long int lli;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<lli> vli;
typedef vector<pii> vii;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> OrderedSet;

const int MOD	= 1e9 + 7;
const double PI = acos(-1.0);

#define fastio			ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define sz(a) int((a).size())
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define all(c) (c).begin(),(c).end()
#define tr(c,i) for(auto &i : c)
#define present(c,x) ((c).find(x) != (c).end())
#define cpresent(c,x) (find(all(c),x) != (c).end())
#define rep(i, n)    for(int i = 0; i < (n); ++i)
#define repA(i, a, n)  for(int i = a; i <= (n); ++i)
#define repD(i, a, n)  for(int i = a; i >= (n); --i)
#define fill(a)  memset(a, 0, sizeof (a))
#define error(x) cerr << #x << " = " << (x) << endl
#define errorp(p) cerr << #p << " = " << (p.first) << ", " << (p.second) << endl

template<typename T> T gcd(T a, T b){return(b?__gcd(a,b):a);}
template <typename T> T lcm(T a, T b){return (a*b)/gcd(a,b); }
int mul(int a, int b, int c){lli res=(lli)a*b;return (int)(res>=c?res%c:res);}
template<typename T>T power(T e, T n, T m){T x=1,p=e;while(n){if(n&1)x=mul(x,p,m);p=mul(p,p,m);n>>=1;}return x;}
int mod_neg(int a, int b, int c){int res;if(abs(a-b)<c)res=a-b;else res=(a-b)%c;return(res<0?res+c:res);}
template<typename T>T extended_euclid(T a, T b, T &x, T &y){T xx=0,yy=1;y=0;x=1;while(b){T q=a/b,t=b;b=a%b;a=t;\
t=xx;xx=x-q*xx;x=t;t=yy;yy=y-q*yy;y=t;}return a;}
template<typename T>T mod_inverse(T a, T n){T x,y,z=0;T d=extended_euclid(a,n,x,y);return(d>1?-1:mod_neg(x,z,n));}
template <class T>  inline void smax(T &x,T y){ x = max((x), (y));}
template <class T>  inline void smin(T &x,T y){ x = min((x), (y));}

int N, M, Q;
vi a, b, c, d;
lli ans;

void dfs(int idx, vi &arr, int last){
    if(idx == N){
        lli tp = 0;
        rep(i, Q){
            if(arr[b[i]] - arr[a[i]] == c[i]){
                tp += d[i];
            }
        }
        smax(ans, tp);
        return;
    }
    repA(j, last, M){
        arr[idx] = j;
        dfs(idx+1, arr, j);
    }
}

int main(){
    
    fastio;
    
    cin >> N  >> M >> Q;
    a.resize(Q);
    b.resize(Q);
    c.resize(Q);
    d.resize(Q);
    rep(i, Q){
        cin >> a[i] >> b[i] >> c[i] >> d[i];
        a[i]--;
        b[i]--;
    }
    ans = 0;

    vi arr(N);
    dfs(0, arr, 1);

    cout << ans << endl;


}