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: Wrong Answer)

#include<iostream>
#include<vector>
#include<utility>
#include<deque>
#include<algorithm>
#include <iomanip>
#include<cmath>
#include<numeric>

using namespace std;
using ll = long long int;
constexpr int INF = 1 << 29;
constexpr ll MOD = ll(1e9 + 7);
int h, w, n;
vector<int> c;


int main(void) {
	cin >> h >> w;
	c = vector<int>(h * w);
	cin >> n;
	int pos = 0;
	for (int i = 0; i < n; i++) {
		int ai, j;
		cin >> ai;
		for (j = pos; j < pos + ai; j++) {
			c[j] = i + 1;
		}
		pos = j;
	}

	for (int i = 0; i < h; i++) {
		for (int j = 0; j < w; j++) {
			if (i % 2 == 0)
				cout << c[i * w + j];
			else
				cout << c[(i + 1) * w - j - 1];
		}
		cout << endl;
	}
	return 0;
}