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: Wrong Answer)

#include <bits/stdc++.h>
#define ll long long int
#define ld long double
const ll MOD = 998244353;
const ll INF = 1e18;
using namespace std;

vector<string> vec_splitter(string s) {
    s += ',';
    vector<string> res;
    while(!s.empty()) {
        res.push_back(s.substr(0, s.find(',')));
        s = s.substr(s.find(',') + 1);
    }
    return res;
}
void debug_out(
vector<string> __attribute__ ((unused)) args,
__attribute__ ((unused)) int idx, 
__attribute__ ((unused)) int LINE_NUM) { cerr << endl; } 
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
    if(idx > 0) cerr << ", "; else cerr << "Line(" << LINE_NUM << ") ";
    stringstream ss; ss << H;
    cerr << args[idx] << " = " << ss.str();
    debug_out(args, idx + 1, LINE_NUM, T...);
}
#ifdef XOX
#define deb(...) debug_out(vec_splitter(#__VA_ARGS__), 0, __LINE__, __VA_ARGS__)
#else
#define deb(...) 42
#endif

    ll dp[20][20][110];
    ll prize[20][20][110];

main(){    
    #ifdef XOX
        freopen("D:\\V S Code\\cpp\\competitiveProgramming\\Input.txt", "r", stdin);
        freopen("D:\\V S Code\\cpp\\competitiveProgramming\\OPT.txt", "w", stdout);
    #endif

    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    
    int n, m, q;
    cin >> n >> m >> q;

    memset(dp, 0, sizeof(dp));
    memset(prize, 0, sizeof(prize));
    ll a, b, c, d;
    while(q--){
        cin >> a >> b >> c >> d;
        prize[a][b-1][c] = max(prize[a][b-1][c], d); 
        dp[a][b-1][c] = max(dp[a][b-1][c], d); 
    }   
    ll ans = 0;

    for(int i = 1; i <= 10; i++){
        for(int j = i; j <= 10; j++){
            for(int k = 0; k <= 100; k++){
                ll &res = dp[i][j][k];
                res= 0ll;
                for(int x = i; x < j; x++)
                for(int l = 0; l <= k; l++){
                   dp[i][j][k] = max({dp[i][j][k], dp[i][x][l] + dp[x + 1][j][k - l]});
                }
                dp[i][j][k] += prize[i][j][k];
                ans = max(ans, dp[i][j][k]);
                if(ans == 120)  deb(i, j, k);
                // deb(i, j, k, dp[i][j][k]);
            }
        }
    }

    cout << ans;
}