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<stdio.h>
#include<math.h>
#include<stdlib.h>
void swap(int *x,int *y){int tmp=*x;*x=*y;*y=tmp;}
int asc_cmp(const void* X,const void* Y){int x=*(int*)X,y=*(int*)Y;if(x<y)return -1;else if(x==y)return 0;else return 1;}

int n, m, q;
int make_indexes(int *a, int n){
    for(int i=n-1;i>=0;i--){
        if(a[i]<m){
            a[i]++;
            for(int j=i+1;j<n;j++){
                a[j]=a[i];
            }
            return 0;
        }
    }
    return 1;
}

int main(){
    scanf("%d%d%d",&n,&m,&q);
    int a[100],b[100],c[100],d[100];
    for(int i=0;i<q;i++){
        scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
    }

    int x[20];for(int i=0;i<n;i++)x[i]=1;
    int flg=0;
    int ans=0;
    int sum =0;
    while(1){
        sum=0;
        //for(int i=0;i<n;i++)printf("%d ",x[i]);
        //printf("\n");
        flg = 0;
        if(!flg){
            for(int i=0;i<q;i++){
                if(x[b[i]-1]-x[a[i]-1]==c[i]){
                    sum+=d[i];
                }
            }
        }
        if(sum>ans){
            ans=sum;
        }
        if (make_indexes(x, n)){
            break;
        }
    }

    printf("%d\n",ans);
}