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>
#include<string.h>
int nxpt(char *s){
  int i,l=strlen(s),max=0,f[260]={0};
  for(l--;l+1&&max<=s[l];l--)f[max=s[l]]++;
  if(l<0)return 0;
  f[s[l]]++;
  for(i=s[l]+1;f[i]==0;i++);
  f[s[l]=i]--;
  for(i=0;s[++l];f[s[l]=i]--){
    while(f[i]==0)i++;
  }
  return 1;
}
int main(){
  int n,m,q,a[100],b[100],c[100],d[100],e[100],i,j,f,r;
  long long ans=0;
  char s[100];
  scanf("%d %d %d",&n,&m,&q);
  for(i=0;i<q;i++)scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
  for(i=0;i<n;i++)s[i]='0';
  for(i=0;i<m;i++)s[i+n]='1';
  s[n+m]=0;
  do{
    for(i=j=r=0;s[i];i++){
      if(s[i]=='1')j++;
      else         e[r++]=j;
    }
    //for(i=0;i<r;i++)printf("%d ",e[i]);printf("\n");
    for(i=f=0;i<q;i++){
      if(e[b[i]-1]-e[a[i]-1]==c[i])f+=d[i];
    }
    if(ans<f)ans=f;
  }while(nxpt(s));
  printf("%lld\n",ans);
  return 0;
}