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>
#include <stdlib.h>
#include <stdint.h> // uint64_t

#define max(a,b) ((a) > (b) ? (a) : (b))
#define min(a,b) ((a) > (b) ? (b) : (a))
#define BUF_SIZE 20
#define DIVISOR 1000000007

int get_int3(int *a1, int *a2, int *a3) {
#ifdef BUF_SIZE
  char line[BUF_SIZE];
  if(!fgets(line, BUF_SIZE, stdin)) return -1;
  sscanf(line, "%d %d %d", a1, a2, a3);
#else
#error
#endif
  return 0;
}

int get_int4(int *a1, int *a2, int *a3, int *a4) {
#ifdef BUF_SIZE
  char line[BUF_SIZE];
  if(!fgets(line, BUF_SIZE, stdin)) return -1;
  sscanf(line, "%d %d %d %d", a1, a2, a3, a4);
#else
#error
#endif
  return 0;
}

struct query {
    int v1;
    int v2;
    int dist;
    int point;
};

#define QUERY_MAX 50
#define NUM_MAX 10

int desc_by_point(const void *a1, const void *a2) {
    struct query *qa1 = (struct query*)a1;
    struct query *qa2 = (struct query*)a2;
    return qa2->point - qa2->dist;
}

struct wufind {
    int parent;
    int depth;
    int dist;
};

struct wufind *create_wufind(int vs) {
    static struct wufind wufind[NUM_MAX];
    int i;
    for(i = 0; i < vs; i++) {
        wufind[i].parent = i;
        wufind[i].depth = 0;
        wufind[i].dist = 0;
    }
    return wufind;
}

int get_root(struct wufind *wu, int v) {
    int base = v;
    struct wufind *elem = &wu[v];
    if(base == elem->parent) {
        return base;
    } else {
        int root = get_root(wu, elem->parent);
        struct wufind *par = &wu[elem->parent];
        elem->dist += par->dist;
        // compression
        elem->parent = root;
        return root;
    }
} 

int get_diff(struct wufind *wufind, int v1, int v2) {
    get_root(wufind, v1);
    get_root(wufind, v2);
    return wufind[v2].dist - wufind[v1].dist;
}

int is_connect(struct wufind *wufind, int v1, int v2) {
    int root1 = get_root(wufind, v1);
    int root2 = get_root(wufind, v2);
    return root1 == root2;
}

// assume that v1, v2 gets compresson done
void unite(struct wufind *wufind, int v1, int v2, int diff) {
    int root1 = wufind[v1].parent;
    int root2 = wufind[v2].parent;
    int w1 = wufind[v1].dist;
    int w2 = wufind[v2].dist;
    struct wufind *wr1 = &wufind[root1];
    struct wufind *wr2 = &wufind[root2];
    // 2->1
    int new_dist = w1 + diff - w2;
    if(wr1->depth >= wr2->depth) {
        if(wr1->depth == wr2->depth) wr1->depth++;
        // 2->1
        wr2->parent = root1;
        wr2->depth = wr1->depth;
        wr2->dist = new_dist;
    } else {
        // 1->2
        wr1->parent = root2;
        wr1->depth = wr2->depth;
        wr1->dist = -new_dist;
    }
    return;
}

int check_limit(struct wufind *wufind, int size, int limit) {
    int i;
    int mi = limit+1;
    int ma = -limit-1;
    for(i = 1; i < size; i++) {
        int res = get_diff(wufind, 0, i);
        mi = min(res, mi);
        ma = max(res, ma);
    }
    return (ma-mi)<=limit;
}


int main(void) {
    int num, limit, qs;
    struct query queries[QUERY_MAX];
    get_int3(&num, &limit, &qs);

    int i, j;
    for(i = 0; i < qs; i++) {
        int v1, v2, dist, point;
        get_int4(&v1, &v2, &dist, &point);
        v1--; v2--;
        queries[i] = (struct query){v1, v2, dist, point};
    }
    qsort(queries, qs, sizeof(struct query), desc_by_point);

    int ans = 0;
    for(i = 0; i < qs; i++) {
        int res = 0;
        struct wufind *wufind = create_wufind(num);
        // weighted union find(rotate)
        for(j = 0; j < qs; j++) {
            int idx = (i+j)%qs;
            struct query q = queries[idx];
            if(!is_connect(wufind, q.v1, q.v2)) {
                if(!check_limit(wufind, num, limit)) continue;
                unite(wufind, q.v1, q.v2, q.dist);
                res += q.point;
            } else {
                if(q.dist == get_diff(wufind, q.v1, q.v2)) {
                    res += q.point;
                }
            }
        }
#ifdef DEBUG
        printf("%d: %d\n", i, res);
#endif
        ans = max(ans, res);
    }
    printf("%d\n", ans);
    return 0;
}