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: Accepted)

#include <iostream>
using namespace std;

int main(void){
    int h, w, n;
    
    cin >> h >> w;
    cin >> n;
    
    int a[10005];
    int field[200][200];

    for(int i=0; i<n; i++){
        cin >> a[i];
    }

    int k=0, cnt=0, l=0;

    for(int i=0; i<n; i++){
        for(int j=0; j<a[i]; j++){
            field[k][l++] = i+1;
            cnt++;
            if(cnt >= w){
                k++;
                l = 0;
                cnt = 0;
            }
        }
    }

    for(int i=1; i<h; i += 2){
        for(int j=0; j<w/2; j++){
            int tmp = field[i][j];
            field[i][j] = field[i][w-1-j];
            field[i][w-1-j] = tmp;
        }
    }

    for(int i=0; i<h; i++){
        for(int j=0; j<w; j++){
            cout << field[i][j];
            if(j != w-1){
                cout << " ";
            }
        }
        cout << endl;
    }

    return 0;
}