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>
using namespace std;

int dp[2][512][256],ara[12][12][12],z=0;


int on(int x,int ps)
{
    return x | (1<<ps);
}


bool ck(int x,int ps)
{
    return (bool)(x & (1<<ps));
}

int main()
{

    int n,m,q,a,b,c,d,i,j,k,mx=0,idx,mask1,mask2,l,fr,sc;
    vector<int>v;

    scanf("%d %d %d",&n,&m,&q);

    for(i=1;i<=q;i++)
    {
        scanf("%d %d %d %d",&a,&b,&c,&d); ara[a][b][c]=d;

        if(d>mx)mx=d;

    }


    if(n==2)printf("%d\n",mx);
    else
    {

        for(idx=n-1;idx>=1;idx--)
        {
            fr=idx%2; sc=(idx+1)%2;

            for(mask1=0;mask1<=511;mask1++)
            {
                for(mask2=0;mask2<=255;mask2++)
                {
                     if(idx==n-1)
                     {

                         int sz=1,t=0,e,f,g,h,r,mx=1,p=0; v.push_back(1);

                         for(i=0;i<=8;i++)
                         {
                             if(ck(mask1,i)==true){  v.push_back(i+2); if(i+2>mx)mx=i+2;  sz++; }
                         }

                         for(i=mx;i<=m;i++)
                         {
                             h=0; t=0;

                             for(j=0;j<idx;j++)
                             {
                                 if(ck(mask2,j)==true)t++;

                                 if(t==sz){ h=-1; break;  }      f=v[t];    h=h+ara[j+1][idx+1][i-f];

                             }

                             p=max(p,h);

                         }

                         dp[fr][mask1][mask2]=p;


                     }
                     else
                     {
                         int sz=1,t=0,e,f,g,h,r,mx=1,p=0; v.push_back(1);

                         for(i=0;i<=8;i++)
                         {
                             if(ck(mask1,i)==true){  v.push_back(i+2); if(i+2>mx)mx=i+2;  sz++; }
                         }

                         for(i=mx;i<=m;i++)
                         {
                             h=0; t=0;

                             for(j=0;j<idx;j++)
                             {
                                 if(ck(mask2,j)==true)t++;

                                 if(t==sz){  h=-1; break; }    f=v[t]; h=h+ara[j+1][idx+1][i-f];

                             }

                             if(h!=-1){
                             e=mask1; g=mask2; if(i>mx){ e=on(mask1,i-2);  g=on(mask2,idx);   }

                             p=max(p,h+dp[sc][e][g]);

                             }

                         }

                         dp[fr][mask1][mask2]=p;

                     }

                     v.clear();

                }

            }

        }


        printf("%d\n",dp[1][0][0]);


    }



    return 0;
}