#include <algorithm>
#include <array>
#include <iostream>
#include <vector>
using std::vector;
int main() {
int n, d, m;
std::cin >> n >> d >> m;
vector<std::array<int, 2>> jobs(m);
for (int i = 0; i < m; i++) {
int day;
std::cin >> day;
jobs[i] = {day, i + 1}; // {request date, index[1...m]}
}
/*
* we sort the jobs by the request date in ascending order
* so that we can test them using isFeasible() in linear time whether they
* can be done in given time using a certain amount of machines
*/
std::sort(begin(jobs), end(jobs));
vector<vector<int>> res;
/** @return true along with schedule if it feasible to finish the jobs otherwise
* false */
auto is_feasible = [&](int machine_count) -> vector<vector<int>> {
vector<vector<int>> schedule(n);
int req_num = 0;
/*
* we simulate from day 1 until the last day n
* we move to the next day if all the machines are used or
* there is no more job requests left on or before this day
*/
for (int day = 1; day <= n; day++) {
for (int j = 0; j < machine_count; j++) {
/*
* if all jobs before and on this day are finished,
* we can go to the next day, even if there are usable machines left
* we can determine that since the vector jobs is sorted
*/
if (jobs[req_num][0] > day) { break; }
/*
* if the current date is before the deadline for the job
* we can add this job to the schedule and move to the next
* job request
*/
if (jobs[req_num][0] + d >= day) {
schedule[day - 1].push_back(jobs[req_num++][1]);
} else { // otherwise, it is not feasible due to deadline
return {};
}
/*
* if we have processed all the requests, we have found a
* feasible solution
*/
if (req_num == m) { return schedule; }
}
}
/*
* if not all the requests can be processed within the given n days,
* then it is not feasible
*/
return {};
};
int lo = 1;
int hi = m;
while (lo < hi) {
int machine_num = (lo + hi) / 2;
/*
* test if the jobs would finish within the deadline
* using the current # of machines i.e., machine_num
*/
vector<vector<int>> curr_result = is_feasible(machine_num);
/*
* if it's possible, we set the right bound as the tested
* machine number and save the current schedule
*/
if (!curr_result.empty()) {
hi = machine_num;
res = curr_result;
} else {
/*
* otherwise, we set the left bound to be the tested number + 1
* and test the next machine_num again
*/
lo = machine_num + 1;
}
}
std::cout << lo << '\n';
for (int i = 0; i < n; i++) {
for (int idx : res[i]) { std::cout << idx << " "; }
std::cout << 0 << '\n';
}
}
