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 <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <cmath>
#include <cstdio>
#include <sstream>

using namespace std;

int main(){
  int h, w, n;
  vector<int> o;
  cin >> h >> w;
  cin >> n;
  int flag=0;
  for (int i=1;i<=n;i++) {
    int a;
    char c;
    cin >> a;
    for (int j=0;j<a;j++) {
      if (o.size()==w) {
	if (flag % 2 == 1) {
	  reverse(o.begin(), o.end());
	}
	for (int k = 0;k<o.size();k++) {
	  cout << o.at(k);
	  if (k<o.size()-1) {
	    cout << " ";
	  } else {
	    cout << endl;
	  }
	}
	o.erase(o.begin(), o.end());
	flag++;
      }
      //      c = to_string(i)[0];
      //      cout << i << "\t" << j << "\t" << c << "\t" << a << endl;
      o.push_back(i);
    }
    if (o.size()==w) {
      if (flag % 2 == 1) {
	reverse(o.begin(), o.end());
      }

      for (int k = 0;k<o.size();k++) {
	cout << o.at(k);
	if (k<o.size()-1) {
	  cout << " ";
	} else {
	  cout << endl;
	}
      }
      o.erase(o.begin(), o.end());
      flag++;
    }
  }
}