#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using db = double;
using ii = pair<int, int>;
using iii = pair<ii, int>;
using pl = pair<ll, ll>;
using ti = tuple<int, int, int>;
using tll = tuple<ll, ll, ll>;
using tld = tuple<ld, ld, ld>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<ii>;
using vpl = vector<pl>;
using vc = vector<char>;
using vs = vector<string>;
#define mp make_pair
#define pb push_back
#define pf push_front
#define pp pop_back
#define ppf pop_front
#define ff first
#define ss second
#define lb lower_bound
#define ub upper_bound
#define sz(v) ((int)v.size())
#define all(x) (x).begin() , (x).end()
#define mt make_tuple
#define FAST ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define parse(A, n) for ( ll K = 0; K < n; K++) { cin >> A[K]; }
#define process(A, n) for( ll K = 1; K <= n; K++) { cin >> A[K]; }
#define out(A, n) { for ( ll K = 0; K < n; K++) { cout <<(K!=0?" ":"")<<A[K]; } cout<<'\n'; }
#define show(x) { bool first=true; for(auto& a:x) { if(first) { first=false; } else { cout<<" "; } cout<<a; } cout<<'\n'; }
#define mem1(memo) memset(memo, -1, sizeof memo);
#define mem0(memo) memset(memo, 0, sizeof(memo);
#define print(k) printf("%.2lf\n", k)
ld dist(ld x, ld y, ld a, ld b) { return sqrt((x-a)*(x-a) + (y-b)*(y-b)); }
ll gcd(ll a , ll b){ return b ? gcd(b , a % b) : a ;}
ll lcm(ll a , ll b){ return (a * b) / gcd(a , b);}
ll fact(ll n) { return n > 1?(n * fact(n-1)):1;}
void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void IO(string s = "") { if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } }
const int N = 124;
const int M = 1e5+24;
const int inf = 1e9;
const int MOD = 1e9+7;
const int nx[4] = {0, 0, 1, -1}, ny[4] = {1, -1, 0, 0}; //East, West, South, North
const int dr[8] = {1,1,0,-1,-1,-1, 0, 1}, dc[8] = {0,1,1, 1, 0,-1,-1,-1}; // S,SE,E,NE,N,NW,W,SW
const long long INF = 1e18;
const double EPS = 1e-9;
const double PI = 3.14159265358979323846;
/** |||||||||||||||||||||||||||||||| SOLUTION |||||||||||||||||||||||||||||||| **/
ll n, d, m;
bool can(ll val, vpl& t)
{
ll maxd=0, endT[val]={0};
for(ll l=0, curr=0;l<m;l++, curr++) {
if(curr==val) curr=0;
if(endT[curr]+1>t[l].ff) {
endT[curr]++;
maxd=max(maxd, endT[curr]-t[l].ff);
}
else endT[curr]=t[l].ff;
}
return maxd<=d;
}
void solve()
{
/// JUST KEEP GOING
cin>>n>>d>>m; vpl t(m);
for(ll l=0;l<m;l++) {
cin>>t[l].ff; t[l].ss=l+1;
}
sort(all(t)); ll lo=0, hi=m; ll ans=LONG_LONG_MAX;
while(lo<hi) {
ll mid=(lo+hi)/2;
if(can(mid, t)) { hi=mid; ans=mid; }
else lo=mid+1;
}
cout<<ans<<'\n';
vl sol[M]; ll endT[ans]{0};
for(ll l=0, curr=0;l<m;l++, curr++) {
if(curr==ans) curr=0;
endT[curr]=max(t[l].ff, endT[curr]+1);
sol[endT[curr]].pb(t[l].ss);
}
for(ll l=1;l<=n;l++) {
for(auto u:sol[l]) cout<<u<<" ";
cout<<"0\n";
}
}
int32_t main()
{
FAST; /// IO("outofplace");
int t; t=1;
while(t--)
{
solve();
}
return 0;
}
/*** Stuff you should look for :
* int overflow, array bounds.
* special cases (n=1?).
* Observe The Constraints.
* do smth instead of nothing and stay organized.
* Reformulate The Problem Into Something More Theoretical.
* Use flush operator when having TLE.
* WRITE STUFF DOWN.
* DON'T GET STUCK ON ONE APPROACH.
Think of:
* Brute force, Greedy(High->Low), Dynamic Programming, Divide and Conquer.
* BFS, DFS, Flood Fill, Topological Sort, Bipartite Check, Articulation Points, SCC.
* Minimum, Maximum Spanning Tree, SSSP, All-Pairs Shortest Paths, Network Flow.
* Disjoint Set Union, Segment tree, Math, Two Pointers, Strings.
Motivation :
* Change Your Approach.
* Don't Spend Too Much Time On The Problem: Move On!
* If You Don't Fight You Cannot Win.
***/
