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 <bits/stdc++.h>
#define x first
#define y second
#define mp make_pair
#define all(a) a.begin(), a.end()
#define pub push_back
#define ll long long
					  
using namespace std;

int n, m, k;
int a[1000000];
int q[111][111];

int main(){
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	//freopen("seating.in", "r", stdin);
	//freopen("seating.out", "w", stdout);
	ios_base::sync_with_stdio(0); cin.tie(0);
	cin >> n >> m >> k;
	for (int i = 0; i < k; i++) cin >> a[i];
	int uk = 0;
	
	for (int i = 0; i < n; i++){
		if (i % 2 == 0){
			for (int j = 0; j < m; j++){
				q[i][j] = uk + 1;
				a[uk]--;
				if (a[uk] == 0) uk++;
			}
		} else {
			for (int j = m - 1; j >= 0; j--){
				q[i][j] = uk + 1;
				a[uk]--;
				if (a[uk] == 0) uk++;
			}
		}
	}

	for (int i = 0; i < n; i++){
		for (int j = 0; j < m; j++){
			cout << q[i][j] << ' ';
		}
		cout << endl;
	}
}