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;

template<typename T>
istream& operator >> (istream& istr, vector<T>& v){
    for(T& x: v) istr >> x;
    return istr;
}

template<typename T, typename U>
istream& operator >> (istream& istr, pair<T,U>& p){
    istr >> p.first >> p.second;
    return istr;
}

template<typename T>
ostream& operator << (ostream& ostr, vector<T>& v){
    if(!v.empty()){
        ostr << v.front();
        for(auto itr = ++v.begin(); itr != v.end(); itr++)
            // ostr << *itr;
            ostr << " " << *itr;
    }
    return ostr;
}

template<typename T, typename U>
ostream& operator << (ostream& ostr, pair<T,U>& p){
    // ostr << p.first << p.second;
    // ostr << p.first << ", " << p.second;
    ostr << "(" << p.first << ", " << p.second << ")";
    return ostr;
}

template<typename T>
ostream& operator << (ostream& ostr, vector<vector<T>>& vv){
    if(!vv.empty()){
        ostr << vv.front();
        for(auto itr = ++vv.begin(); itr != vv.end(); itr++)
            ostr << endl << *itr;
    }
    return ostr;
}

template<typename T, typename U>
ostream& operator << (ostream& ostr, vector<pair<T,U>>& vp){
    if(!vp.empty()){
        ostr << vp.front();
        for(auto itr = ++vp.begin(); itr != vp.end(); itr++)
            ostr << endl << *itr;
    }
    return ostr;
}

typedef long long ll;
typedef vector<ll> vll;
typedef vector<char> vc;
typedef vector<double> vd;
// vll v(n); v[i]=x;
typedef vector<vll> vvll;
typedef vector<vc> vvc;
typedef vector<vd> vvd;
// vvll v(m,vll(n)); v[i][j]=x;
typedef pair<ll,ll> pll;
typedef pair<double,double> pd;
typedef vector<pll> vpll;
typedef vector<pd> vpd;
// vpll v(n); v[i]=make_pair(x,y); v[i].first/second;
// v.front() : v[0]
// v.back()  : v[v.size()-1]

#define PI 3.1415926535897932
#define EPS 1e-12
#define INF ((ll)1e+12)
#define REP(i,n) for(ll i=0; i<(n); i++)
#define RREP(i,n) for(ll i=(n)-1; i>=0; i--)
#define FOR(i,a,b) for(ll i=(a); i<=(b); i++)
#define FORR(i,a,b) for(ll i=(b); i>=(a); i--)
#define fix(x) cout << fixed << setprecision(x)
#define dump(x) cout << #x << " = " << (x) << endl
#define all(x) (x).begin(),(x).end()

template<typename T> T sum(vector<T>& v){return accumulate(all(v), (T)0);}
template<typename T> inline void sort(vector<T>& v){sort(all(v));}
template<typename T> inline void rsort(vector<T>& v){sort(all(v), greater<T>());}
template<typename T> inline void chmin(T& a, T b){if(a>b) a=b;}
template<typename T> inline void chmax(T& a, T b){if(a<b) a=b;}

int main(){
    ll n, m, q, ans = 0;
    cin >> n >> m >> q;

    vvll x(q, vll(4));
    REP(i,q) REP(j,4) cin >> x[i][j];
    REP(i,q){
        x[i][0]--;
        x[i][1]--;
    }
    sort(x);

    vll a(n, 1);
    while(1){
        ll tmp = 0;
        REP(i,q){
            if(a[x[i][1]]-a[x[i][0]]==x[i][2]) tmp += x[i][3];
            // dump(tmp);
            chmax(ans, tmp);
        }
        RREP(i,n){
            if(a[i]!=m){
                a[i]++;
                FOR(j,i+1,n-1) a[j]=a[i];
                break;
            }
        }
        // dump(a);
        if(a[0]==m) break;
    }
    cout << ans << endl;
    return 0;
}