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>

using namespace std;
typedef long long ll;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const ll LINF = 1e18;

int main() {

    int h, w; cin >> h >> w;

    int n; cin >> n;

    vector<pair<int, int>> p(n);
    for (int i = 0; i < n; ++i) {
        int a; cin >> a;
        p[i] = make_pair(a, i+1);
    }

    sort(p.rbegin(), p.rend());

    vector<int> a;

    for (auto t : p) {
        for (int i = 0; i < t.first; ++i) {
            a.push_back(t.second);
        }
    }

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

        cout << endl;
    }
}