Problem:
Score : 400 points

Problem Statement

We have a grid with H rows and W columns of squares.
Snuke is painting these squares in colors 1, 2, ..., N.
Here, the following conditions should be satisfied:

For each i (1 ≤ i ≤ N), there are exactly a_i squares painted in Color i. Here, a_1 + a_2 + ... + a_N = H W.

For each i (1 ≤ i ≤ N), the squares painted in Color i are 4-connected. That is, every square painted in Color i can be reached from every square painted in Color i by repeatedly traveling to a horizontally or vertically adjacent square painted in Color i.

Find a way to paint the squares so that the conditions are satisfied.
It can be shown that a solution always exists.

Constraints

1 ≤ H, W ≤ 100

1 ≤ N ≤ H W

a_i ≥ 1

a_1 + a_2 + ... + a_N = H W

Input

Input is given from Standard Input in the following format:

H W
N
a_1 a_2 ... a_N

Output

Print one way to paint the squares that satisfies the conditions.
Output in the following format:

c_{1 1} ... c_{1 W}
:
c_{H 1} ... c_{H W}

Here, c_{i j} is the color of the square at the i-th row from the top and j-th column from the left.

Sample Input 1

2 2
3
2 1 1

Sample Output 1

1 1
2 3

Below is an example of an invalid solution:

1 2
3 1

This is because the squares painted in Color 1 are not 4-connected.

Sample Input 2

3 5
5
1 2 3 4 5

Sample Output 2

1 4 4 4 3
2 5 4 5 3
2 5 5 5 3

Sample Input 3

1 1
1
1

Sample Output 3

1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;
#define FAST ios_base::sync_with_stdio(false); cin.tie(0);
#define LLINF (long long) 1e18//1234567890987654321
#define INF 1234567890
#define pb push_back
#define ins insert
#define f first
#define s second	
#define db 0
#define EPS (1e-7)    //0.0000001 the value
#define PI (acos(-1))
#define MAXN 300006
#define MAXK 26
#define MAXX 15000006
#define ll long long int
#define ld long double
#define rep0(kk, l1, l2)for(ll kk = l1; kk < l2; kk++)
#define rep1(kk, l1, l2)for(ll kk = l1; kk <= l2; kk++)
#define foritr(itr, A) for(set<ll>::iterator itr = A.begin(); itr != A.end(); itr++)
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());    //can be used by calling rng() or shuffle(A, A+n, rng)
#define FOR(ii, ss, ee) for(ll ii = ss; ii < ee; ii++)
#define cr(x) cerr << #x << " = " << x << "\n";
#define crA(x, A) cerr << #x << " = " << A[x] << "\n";
#define spacing cout << " ";
#define mmst(x, v) memset((x), v, sizeof ((x)));
#define bg(ms) (*ms.begin())
#define ed(ms) (*prev(ms.end(), 1))
#define addedge(a, b, c, v) v[(a)].pb(pi((b), (c))); v[(b)].pb(pi((a), (c)))
#define ph push
#define btinpct(x) __builtin_popcountll(x)
#define p2(x) (1LL<<(x))
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
typedef pair <ll, ll> pi;
typedef pair <ll, pi> spi;
typedef pair <pi, pi> dpi;
inline ll rand(ll x, ll y) { ++y; return (rng() % (y-x)) + x; } //inclusivesss
ll n, h, w, co_h, co_w, f=1;
pi colour[MAXN];
ll grid[106][106];
int main()
{
	cin >> h >> w >> n;
	FOR(i,0,n)
	{
		cin >> colour[i].f; colour[i].s = i;
	}
	mmst(grid, -1);
	sort(colour, colour+n, greater<pi>());
	for(ll i = 0; i < n; i++)
	{
		ll ind = colour[i].s;
		while(colour[i].f --)
		{
			grid[co_h][co_w] = ind;
          	co_h += f;
			if(co_h >= h || co_h < 0) { co_w++; f = (f == 1 ? -1 : 1); co_h += f; }
		}
	}
	FOR(i,0,h)
	{
		FOR(j,0,w)
		{
          if(grid[i][j] == -1)assert(0);
			cout << grid[i][j] + 1 << " ";
		}
		cout << "\n";
	}
}