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: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <vector>

using namespace std;

int main()
{
    int H, W, N;
    if (scanf("%d%d%d", &H, &W, &N) < 3) return 0;

    int rnum = 0;
    int cnum = 0;

    int a;

    bool flg = true;

    vector<int> vi(W);

    for (int i = 0; i < N; i++)
    {
        if (scanf("%d", &a) < 1) return 0;

        int bk;

        int j;

        for (j = cnum; j < cnum + a; j++)
        {
            if (W <= j)
            {
                bk = j - cnum + 1;
                cnum = 0;
                //flg = !flg;
                break;
            }
            vi[j] = i + 1;
        }

        if (j == W)
        {
            if (flg)
            {
                for (int k = 0; k < W; k++)
                {
                    if (k == 0) printf("%d", vi[k]);
                    else printf(" %d", vi[k]);
                }
                printf("\n");
            }
            else
            {
                for (int k = W - 1; 0 <= k; k--)
                {
                    if (k == W - 1) printf("%d", vi[k]);
                    else printf(" %d", vi[k]);
                }
                printf("\n");
            }
            flg = !flg;
        }
        else
        {
            cnum = j;
        }
    }

    return 0;
}