#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
vi todo[MX], ans[MX];
int N, D, M;
int gen(int x, bool b = 0) {
// return 0;
queue<pi> cur;
int mx = 0;
FOR(i,1,N+1) {
for (int j: todo[i]) cur.push({j,i});
for (int j = 0; j < x && sz(cur) > 0; ++j) {
if (b) ans[i].pb(cur.front().f);
mx = max(mx,i-cur.front().s);
cur.pop();
}
}
if (sz(cur)) return MOD;
return mx;
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N >> D >> M;
FOR(i,1,M+1) {
int x; cin >> x;
todo[x].pb(i);
}
int lo = 1, hi = 1000000;
while (lo < hi) {
int mid = (lo+hi)/2;
if (gen(mid) <= D) hi = mid;
else lo = mid+1;
}
gen(lo,1);
cout << lo << "\n";
FOR(i,1,N+1) {
for (int j: ans[i]) cout << j << " ";
cout << "0\n";
}
}
// read the question correctly (is y a vowel? what are the exact constraints?)
// look out for SPECIAL CASES (n=1?) and overflow (ll vs int?)
