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 y=1,i=1,j=1,k=0,m,n,sum=0,x,z;
    scanf("%d%d",&m,&n);
    int a[110][110]={0};
    scanf("%d",&z);
    while(sum<m*n){
        scanf("%d",&x);
        sum+=x;
        i=1;
            do{
            while(i<=x&&k<n&&a[j][k+1]==0){
                a[j][++k]=y;
                i++;
            }
            while(i<=x&&j<m&&a[j+1][k]==0){
                a[++j][k]=y;
                i++;
            }
            while(i<=x&&k>1&&a[j][k-1]==0){
                a[j][--k]=y;
                i++;
            }
            while(i<=x&&j>1&&a[j-1][k]==0){
                a[--j][k]=y;
                i++;
            }}while(i<=x);
        y++;
    }
    j=1;k=1;
    for(i=1;i<=m*n;i++){
        if(i%n==0) {
            printf("%d ",a[j][k]);
            j++;
            k=1;
            printf("\n");
        }
        else printf("%d ",a[j][k++]);
    }
    return 0;
}