#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
int n, d, m;
vector<int> a; // availability of each job (1‑based)
vector<vi> answer; // answer[day] = list of jobs on that day
/*------------------------------------------------------------
Check if we can schedule all jobs with at most `machines`
parallel jobs per day.
If `fill` is true, the global vector `answer` is filled with
a concrete schedule.
------------------------------------------------------------*/
bool feasible(int machines, bool fill = false) {
// release[day] = ids of jobs that become available on this day
vector<vector<int>> release(n + 2);
for (int i = 0; i < m; ++i) {
if (a[i] > n) return false; // impossible input
release[a[i]].push_back(i);
}
// (deadline, job_id) – smallest deadline on top
using pii = pair<int,int>;
priority_queue<pii, vector<pii>, greater<pii>> pq;
if (fill) answer.assign(n + 1, {});
for (int day = 1; day <= n; ++day) {
// new jobs become available
for (int id : release[day]) {
int deadline = a[id] + d;
if (deadline > n) deadline = n; // days after n are useless
pq.emplace(deadline, id);
}
int slots = machines;
while (slots && !pq.empty()) {
auto [deadline, id] = pq.top(); pq.pop();
if (deadline < day) return false; // missed its last possible day
if (fill) answer[day].push_back(id + 1); // store 1‑based id
--slots;
}
}
// after day n there must be no remaining jobs
return pq.empty();
}
/*------------------------------------------------------------*/
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
if(!(cin >> n >> d >> m)) return 0;
a.resize(m);
for (int i = 0; i < m; ++i) cin >> a[i];
int lo = 1, hi = m, best = m;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (feasible(mid)) {
best = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
// build the concrete schedule for the optimal number of machines
feasible(best, true);
cout << best << '\n';
for (int day = 1; day <= n; ++day) {
for (int id : answer[day]) cout << id << ' ';
cout << "0\n";
}
return 0;
}
