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 <stdio.h>
int main(void){
    //変数定義
    int N;//N文字
    int M;//M種類
    int Q;//フィルターの数
    int point = 0;//現在のポイント
    int maxpoint = 0;//最大のポイント
    scanf("%d %d %d",&N,&M,&Q);//１行目入力
    int f[Q][4];//フィルター格納用配列、0~(Q-1)
    int l[N];//数列格納用配列、0~(N-1)
    //入力プログラム
    for(int i=0;i<Q;i++){//Q回繰り返す
    	scanf("%d %d %d %d",&f[i][0],&f[i][1],&f[i][2],&f[i][3]);//１行読み取り
    }
    //ループ＆数列作成
    //N文字の数列
    //初期化
    for(int i=1;i<=N;i++){
    	l[i-1]=i;
    }
    
    while(l[0]!=M-N+1){
        //printf("%d %d %d\n",l[0],l[1],l[2]);
    	//点数計算処理
    	point = 0;
    	for(int i=0;i<Q;i++){//Q回繰り返す
    	    //printf("%d %d %d %d",f[i][0],f[i][1],f[i][2],f[i][3]);
    		if(l[f[i][1]-1]-l[f[i][0]-1]==f[i][2]){
    		    //printf("*\n");
    			point += f[i][3];
    		}else{
    		    //printf("\n");
    		}
    	}
    	if(maxpoint<=point)
          	maxpoint = point;
       	
    	//新規数列作成処理
    	//各n文字目について処理
    	for(int i=N-1;i>=0;i--){
    		if(i==N-1){
    			if(l[i]<M){//右端の数がMより小さいなら
    			l[i]++;
    			break;
    			}
    		}else{
    			if(l[i]<l[i+1]){//右端の数がMより小さいなら
           			l[i]++;
           			break;
           		}
           	}
    	}
    }
    for(int i=0;i<Q;i++){//Q回繰り返す
    	    //printf("%d %d %d %d",f[i][0],f[i][1],f[i][2],f[i][3]);
    		if(l[f[i][1]-1]-l[f[i][0]-1]==f[i][2]){
    		    //printf("*\n");
    			point += f[i][3];
    		}else{
    		    //printf("\n");
    		}
    	}
    	if(maxpoint<=point)
          	maxpoint = point;
       	
    printf("%d",maxpoint);
}