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 (Clang 3.8.0) (judge verdict: Accepted)

#include<stdio.h>  
   
int a[105][105],b[10000+50];  
   
int main ()  
{  
    int h,w;  
    while(scanf("%d %d",&h,&w)!=EOF)  
    {  
        int i,j,t,k=0;  
        scanf("%d",&t);  
        for(i=0;i<t;i++)  
            scanf("%d",&b[i]);  
        for(i=0;i<h;i++)  
        {  
            if(i%2==0)  
                for(j=0;j<w;j++)  
                {  
                    a[i][j]=k+1;  
                    b[k]--;  
                    if(b[k]==0)  
                        k++;  
                }  
            if(i%2==1)  
                for(j=w-1;j>=0;j--)  
                {  
                    a[i][j]=k+1;  
                    b[k]--;  
                    if(b[k]==0)  
                        k++;  
                }  
        }  
        for(i=0;i<h;i++)  
        {  
            for(j=0;j<w-1;j++)  
                printf("%d ",a[i][j]);  
            printf("%d\n",a[i][w-1]);  
        }  
    }  
}