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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <queue>
#include <vector>
#include <cmath>
#define int long long
typedef long long ll;
using namespace std;

ll N, M, Q, A[50], B[50], C[50], D[50], maxpoint=0;

void check(ll x){
    ll maxcheck = 0;
    for(int i=0;i<Q;i++){
        ll numa = x/pow(10, A[i]-1);
        ll numb = x/pow(10, B[i]-1);
        if(C[i]==numa%10-numb%10) maxcheck += D[i];
    }
    cout << x << " " << maxcheck << endl;
    maxpoint = max(maxpoint, maxcheck);
    return;
}

void dfs(ll x){
    if(x>pow(10,N-1)) {
        check(x);
        return;
    }
    else if(x%10<=M&&x<pow(10,N-1)){
        x = 10*x + x%10;
        if(x==0) for(int i=1;i<=M-x%10;i++) dfs(x+i);
        else for(int i=0;i<=M-x%10;i++) dfs(x+i);
    }
    return;
}

signed main(){
    cin >> N >> M >> Q;
    for(int i=0;i<Q;i++) cin >> A[i] >> B[i] >> C[i] >> D[i];
    dfs(0);
    cout << maxpoint << endl;
    return 0;
}