#include <bits/stdc++.h>
using namespace std;
// Using int is faster than long long and sufficient for N, M <= 2e9 indices
// We only use long long if calculating sums (not needed here)
typedef int num;
// Global array for the check function to avoid re-allocation overhead
// Assuming max M is around 2e5, usually safe to allocate slightly more
const int MAX_M = 200005;
num finish_times[MAX_M];
int main() {
// 1. Ultra-fast I/O
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
num n, d, m;
if (!(cin >> n >> d >> m)) return 0;
// 2. Store tasks.
// We strictly need pairs to track original indices for the final output.
vector<pair<num, num>> v(m);
for(int i = 0; i < m; i++) {
cin >> v[i].first;
v[i].second = i;
}
sort(v.begin(), v.end());
// 3. Binary Search
// Lower bound 1, Upper bound M (worst case: all tasks on one day needs M capacity)
num ans = m;
num low = 1, high = m;
while(low <= high) {
num k = (low + high) / 2;
bool possible = true;
// --- THE OPTIMIZATION ---
// Instead of simulating days, use the recurrence relation:
// Task 'i' reuses the slot of task 'i-k'.
// Complexity: Strict O(M) with minimal operations.
for(int i = 0; i < m; i++) {
num arrival = v[i].first;
// If i < k, it's the first task in its "thread", so it finishes at 'arrival'.
// If i >= k, it starts after the previous task in this thread finishes (finish_times[i-k] + 1)
num prev_finish = (i >= k) ? finish_times[i-k] : 0;
// The task finishes at the later of: when it arrives, or when the slot opens up
num current_finish = max(arrival, prev_finish + 1);
// Check constraints immediately
if(current_finish > n || current_finish - arrival > d) {
possible = false;
break;
}
finish_times[i] = current_finish;
}
// ------------------------
if(possible) {
ans = k;
high = k - 1;
} else {
low = k + 1;
}
}
cout << ans << "\n";
// 4. Output Logic
// The greedy simulation is optimal for printing and runs in O(N + M)
// Since we only print once, this is acceptable.
num task_idx = 0;
for(num day = 1; day <= n; day++) {
num capacity = ans;
// While tasks are available (arrived <= today) and we have capacity
while(capacity > 0 && task_idx < m && v[task_idx].first <= day) {
cout << v[task_idx].second + 1 << " ";
task_idx++;
capacity--;
}
cout << "0\n";
}
return 0;
}
