#include <bits/stdc++.h>
using namespace std;
#define vi vector<int>
#define ii pair<int,int>
#define vii vector<ii>
#define ALL(x) begin(x), end(x)
void solve() {
int n, d, m;
cin >> n >> d >> m;
// jobs[i] = {availability_day, job_id}
vii jobs(m);
for (int i = 0; i < m; ++i) {
cin >> jobs[i].first;
jobs[i].second = i + 1; // 1-based job id for printing
}
sort(ALL(jobs)); // by availability then id
// dist[day] = list of job ids scheduled on that day (1..n)
vector<vi> dist(n + 1);
auto feasible_and_fill = [&](int machines) -> bool {
for (int i = 0; i <= n; ++i) dist[i].clear();
// multiset of "next free day" for each machine. Initially all are free on day 1 (i.e., next_free = 1)
multiset<int> next_free;
for (int i = 0; i < machines; ++i) next_free.insert(1);
for (auto [a, id] : jobs) {
// find a machine with next_free <= a + d
auto it = next_free.upper_bound(a + d - 1);
if (it == next_free.begin()) return false; // all machines next_free > a + d
--it; // machine with the largest next_free <= a + d
int t = *it; // earliest day this machine can start
int s = max(t, a); // actual start day
if (s > n) return false; // out of horizon
// s is guaranteed <= a + d by the upper_bound choice
// schedule on day s
dist[s].push_back(id);
// update machine: next free becomes s + 1
next_free.erase(it);
next_free.insert(s + 1);
}
return true;
};
int l = 1, r = min(n, m), ans = m;
while (l <= r) {
int mid = (l + r) / 2;
if (feasible_and_fill(mid)) {
ans = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
// Recompute schedule for the minimal ans so dist matches what we print
feasible_and_fill(ans);
cout << ans << '\n';
for (int day = 1; day <= n; ++day) {
for (int id : dist[day]) cout << id << ' ';
cout << "0\n";
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int tt = 1;
while (tt--) solve();
}
