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<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
#include<functional>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<numeric>
#include<limits>
#include<iterator>

#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()

using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ll, char> plc;

int field[110][110];

int main()
{
	int h, w, n;
	cin >> h >> w >> n;

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

	int j = 0,num = 1,k=0,add = 1;
	for (int i = 0; i < h; i++) {
		while (1) {
			field[i][j] = num;
			a[k]--;
			if (a[k] <= 0) {
				k++;
				num++;
			}
			j += add;
			if (j >= w || j < 0) {
				if(j >= w)j = w - 1;
				else j = 0;
				add *= -1;
				break;
			}
		}
	}

	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; 
}