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)

#define _USE_MATH_DEFINES
#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <utility>
#include <complex>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <deque>
#include <tuple>
#include <bitset>
#include <limits>
#include <algorithm>
#include <array>
#include <random>
#include <complex>
#include <regex>
using namespace std;
typedef long double ld;
typedef long long ll;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef complex<ld> compd;
#define quickIO()	{cin.tie(0);	cout.sync_with_stdio(false);}
#define reach(i,a)	for(auto i:a)
#define rep(i,n)	for(int i=0;i<(int)n;i++)
#define REP(i,n)	for(int i=0;i<=(int)n;i++)
#define srep(i,a,n)	for(int i=a;i<(int)n;i++)
#define SREP(i,a,n)	for(int i=a;i<=(int)n;i++)
#define rrep(i,n)	for(int i=n-1;i>=0;i--)
#define RREP(i,n)	for(int i=n;i>=0;i--)
#define all(a)	(a).begin(),(a).end()
#define mp(a,b)	make_pair(a,b)
#define mt	make_tuple
#define pb	push_back
#define fst	first
#define scn second
int bitcnt(ll x) {
	x = ((x & 0xAAAAAAAAAAAAAAAA) >> 1) + (x & 0x5555555555555555);
	x = ((x & 0xCCCCCCCCCCCCCCCC) >> 2) + (x & 0x3333333333333333);
	x = ((x & 0xF0F0F0F0F0F0F0F0) >> 4) + (x & 0x0F0F0F0F0F0F0F0F);
	x = ((x & 0xFF00FF00FF00FF00) >> 8) + (x & 0x00FF00FF00FF00FF);
	x = ((x & 0xFFFF0000FFFF0000) >> 16) + (x & 0x0000FFFF0000FFFF);
	x = ((x & 0xFFFFFFFF00000000) >> 32) + (x & 0x00000000FFFFFFFF);
	return x;
}
int bitcnt(int x) {
	x = ((x & 0xAAAAAAAA) >> 1) + (x & 0x55555555);
	x = ((x & 0xCCCCCCCC) >> 2) + (x & 0x33333333);
	x = ((x & 0xF0F0F0F0) >> 4) + (x & 0x0F0F0F0F);
	x = ((x & 0xFF00FF00) >> 8) + (x & 0x00FF00FF);
	x = ((x & 0xFFFF0000) >> 16) + (x & 0x0000FFFF);
	return x;
}
ll sqrt(ll x) {
	ll left = 0, right = x;
	rep(i, 100) {
		ll mid = (left + right) >> 1;
		if (mid*mid <= x)	left = mid;
		else	right = mid;
	}
	return left;
}
ll gcd(ll a, ll b) {
	return a%b == 0 ? b : gcd(b, a%b);
}
#define debug(x)	printf("Case #%d: ", x)
#define DEBUG 0
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const ld eps = 1e-15;
const int dx[] = { 1,0,-1,0,0 };
const int dy[] = { 0,1,0,-1,0 };

// めんどくさいので、色ごとにまとめて塗りたいですよね。
// これが上手くいく方法を考えると、以下の方法で上手くいきます。
/*
→→→→→↓
↓←←←←←
→→→→→↓
…
*/
// これなら色毎に全部連結になりますよね？（他別解は多数存在すると思います）
// 処理はyの偶奇で上手く分けます。コード読めばわかるかと。

int main() {
	// データの入手
	int h, w;	cin >> h >> w;
	int n;	cin >> n;
	vint a(n);
	rep(i, n)	cin >> a[i];

	// 答えは2次元配列に取り敢えず入れます。
	vector<vint> ret(h, vint(w, 0));
	// 色ごとにループを回して、座標はループの外側に用意しておいて、制御します。
	int x = 0, y = 0;
	rep(i, n) {
		rep(j, a[i]) {
			ret[y][x] = i + 1;
			// yが奇数の時（左へ動かしていきます）
			if (y & 1) {
				// x==0ならyを進め、そうでないなら左へ。
				if (!x)	y++;
				else	x--;
			}
			// yが偶数の時（右へ）
			else {
				// 同様に。
				if (x == w - 1)	y++;
				else	x++;
			}
		}
	}

	// 答えの出力。
	rep(i, h) {
		// j==0の時だけ半角スペースの出力を省略してます
		rep(j, w)	cout << (j ? " " : "") << ret[i][j];
		cout << endl;
	}
	return 0;
}