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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdint.h>
#include<stdbool.h>
#include<tgmath.h>
#include<sys/queue.h>
#include<time.h>
#define ll long long
#define min(a, b) ((a) > (b) ? (b) : (a))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define miin(a, b, c) min((a), min((b), (c)))
#define maax(a, b, c) max((a), max((b), (c)))
#define mod 1000000007

int h, w, n, a[11111], b[111][111];

main(){
    scanf("%d%d%d", &h, &w, &n);
    for(int i=0; i<n; i++){
        scanf("%d", &a[i]);
    }
    int t = 0, u = 0;
    for(int i=0; i<h; i++){
        for(int j=0; j<w; j++){
            int v = u?j:w-1-j;
            b[i][v] = t+1;
            a[t]--;
            if(a[t] == 0){
                t++;
            }
        }
        u ^= 1;
    }
    for(int i=0; i<h; i++){
        for(int j=0; j<w; j++){
            if(j)putchar(32);
            printf("%d", b[i][j]);
        }
        putchar(10);
    }
    return 0;
}