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() {

    int h, w, n, a[10000];
    cin >> h >> w >> n;
    for(int i = 0; i < n; i++) {
        cin >> a[i];
    }

    bool turn = true;
    int x = 1, y = 1, incx = 1, incy = 1;
    int ans[101][101] = {{0}};
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < a[i]; j++) {
            ans[y][x] = i + 1;
            if(turn) x += incx;
            else     y += incy;
            if(turn && ((incx > 0 && x > w) || (incx < 0 && x == 0) || ans[y][x])) {
                x -= incx;
                incx *= -1;
                y += incy;
                turn = !turn;
            }
            if(!turn && ((incy > 0 && y > h) || (incy < 0 && y == 0) || ans[y][x])) {
                y -= incy;
                incy *= -1;
                x += incx;
                turn = !turn;
            }
        }
    }

    for(int i = 1; i <= h; i++) {
        cout << ans[i][1];
        for(int j = 2; j <= w; j++) {
            cout << " " << ans[i][j];
        }
        cout << endl;
    }
}