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 Ruby (2.3.3) (judge verdict: Accepted)

# Here your code !
h, w = gets.chomp.split.map &:to_i
n = gets.chomp.to_i
a = gets.chomp.split.map &:to_i
ans = Array.new(h){Array.new(w,0)}
f = 0
y = x = 0
table = []
a.each.with_index(1) do |d, i|
  d.times{table.push(i)} 
end
i = 0
(h * w * 2).times do
  if(x >= w)
    x = w - 1
    f = 1
  elsif(y >= h)
    y = h - 1
    f = 2
  elsif(x < 0)
    x = 0
    f = 3
  elsif(y < 0)
    y = 0
    f = 0
  elsif(ans[y][x] != 0)
    case f
    when 0 then
      x -= 1
    when 1 then
      y -= 1
    when 2 then
      x += 1
    when 3 then
      y += 1
    end
    f += 1
    f %= 4
  else
    if(ans[y][x] == 0)
      ans[y][x] = table[i] 
      i += 1
    end
  end
  case f
  when 0 then
    x += 1
  when 1 then
    y += 1
  when 2 then
    x -= 1
  when 3 then
    y -= 1
  end
end
ans.each{|aa|puts aa.join(" ")}