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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
int main()
{   
	int H, W, N,i,j,x,y,n=0,sum=0;
	int a[10000], b[10000] = {0}, HW[100][100] = { 0 };
	while ((scanf("%d%d",&H,&W))==2)
	{
		scanf("%d", &N);
		for (i = 0; i < N; i++)
			scanf("%d", &a[i]);
		for (i = 0; i < N; i++)
		{
			for (j = 0; j < a[i]; j++)
				b[n++] = i+1;
		}
		n = 1; HW[x = 0][y = W - 1] = 1;
		while (b[n])
		{
			while (x + 1 < H && !HW[x + 1][y])
				HW[++x][y] = b[n++];
			while (y - 1 >= 0 && !HW[x][y - 1])
				HW[x][--y] = b[n++];
			while (x - 1 >= 0 && !HW[x - 1][y])
				HW[--x][y] = b[n++];
			while (y + 1 < n && !HW[x][y + 1])
				HW[x][++y] = b[n++];
		}
		for (i = 0; i < H; i++)
		{
			for (j = 0; j < W; j++)
			{
				printf("%d ", HW[i][j]);
			}
			printf("\n");
		}
	}
	return 0;
}