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>
#define rep(i, start, end) for (long long i = start; i < end; ++i)
#define repreverse(i, start, end) for (long long i = start; i >= end; --i)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define len(x) ((long long)(x).size())
#define lcm(a, b) ((a) / __gcd((a), (b)) * (b))
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vllvll = vector<vll>;
using vc = vector<char>;
using pll = pair<ll, ll>;
template<class T> void print1d(T x,ll n=-1){if(n==-1)n=x.size();rep(i,0,n){cout<<x[i]<<' ';}cout<<'\n';}
template<class T> void print2d(T x,ll r=-1,ll c=-1){if(r==-1)r=x.size();if(c==-1)c=x[0].size();rep(i,0,r)print1d(x[i],c);}
template<class T, class U> bool haskey(T mp, U key) { return mp.find(key) != mp.end(); }
template<class T, class U> bool isin(T el, U container) { return find(all(container), el) != container.end(); }
template<class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
template<class T> bool even(T n) { return ! (n & 1); }
template<class T> bool odd(T n) { return n & 1; }
template<class T> ll rup(T a, T b) { return a % b ? a / b + 1 : a / b; }
template<class T> ld deg2rad(T deg) { return M_PI * deg / 180.0; }
template<class T> ld rad2deg(T rad) { return 180.0 * rad / M_PI; }
const long double pi = M_PI;
const long long big = 1LL << 50;
const long long inf = 1LL << 60;
const long long mod = 1e9 + 7;
ll intpow(ll a, ll n, ll _mod=numeric_limits<ll>::max()) { ll p=1; while (n) { if (n&1) p=p*a%_mod; a=a*a%_mod; n>>=1; } return p; } 
ll modc(ll a, char op, ll b, ll _mod=mod)
{
        a %= _mod; b %= _mod; ll res = 1;
        switch (op) {
        case '+': res = (a + b) % _mod; break;
        case '-': res = (a - b) % _mod; break;
        case '*': res = a * b % _mod; break;
        case '/': res = modc(a, '*', modc(b, '^', _mod-2, _mod), _mod); break;
        case '^': res = intpow(a, b, _mod); break;
        case 'P': rep(i, a-b+1, a+1) res = modc(res, '*', i, _mod); break;
        case 'C': res = modc(modc(a, 'P', b, _mod), '/', modc(b, 'P', b, _mod)); break;
        } 
        if (res < 0) { res += _mod; } return res;
}

vllvll ans;
ll N, M, Q;

void rec(vll v, ll a, ll cur)
{
    if (a > M) return;
    if (cur == N) {
        ans.push_back(v);
        return;
    }

    v.push_back(a);

    rep(i, 0, M+1) {
        rec(v, a+i, cur+1);
    }
}

int main()
{
    cin >> N >> M >> Q;
    vll tmp;
    rep(i, 1, M+1) rec(tmp, i, 0);    

    vll a(Q), b(Q), c(Q), d(Q);
    rep(i, 0, Q) {
        cin >> a[i] >> b[i] >> c[i] >> d[i];
    }

    ll mx = -inf;
    rep(i, 0, len(ans)) {
        ll sc = 0;
        rep(j, 0, Q)
            if (ans[i][b[j]-1] - ans[i][a[j]-1] == c[j]) {
                sc += d[j];
            }

        chmax(mx, sc);
    }

    cout << mx << endl;
}