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 <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <cstring>
#include <queue>
#include <stack>
#include <functional>
#include <math.h>
#include <iomanip>
using namespace std;
#define Would
#define you
const int INF = 999999999;
const int MOD = 1e9 + 7;

int main() {
	int a, b, c, k[10005];
	cin >> a >> b >> c;
	for (int i = 0; i < c; ++i) {
		cin >> k[i];
	}
	vector<int>v;
	for (int i = 1; i <= c; ++i) {
		for (int j = 0; j < k[i - 1]; ++j) {
			v.push_back(i);
		}
	}
	for (int i = 0; i < a; ++i) {
		if (i % 2 == 0) {
			for (int j = b * i; j < b*i + b; ++j) { cout << v[j]; }
		}
		else {
			for (int j = b * i + b; j >= b * i; --j) { cout << v[j]; }
		}
		cout << endl;
	}
}