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 <bits/stdc++.h>
using namespace std;

int dx[]{0, -1, 0, 1};
int dy[]{1, 0, -1, 0};
int field[1001][1001] = {0};

int main() {
    int h, w;
    cin >> h >> w;
    int n;
    cin >> n;
    vector<int> v(n);
    for(int i=0; i<n; ++i) cin >> v[i];

    for(int i=0; i<n; ++i) {
        int f = 0;
        for(int j=0; j<h; ++j) {
            if(f!=0) break;
            for(int k=0; k<w; ++k) {
                if(field[j][k]==0) {
                    int a = i+1;
                    int a_num = v[i];
                    int c = 0;
                    stack<pair<int, int>> que;
                    que.push(make_pair(j, k));
                    while(!que.empty()) {
                        auto p = que.top();
                        que.pop();
                        int x = p.first;
                        int y = p.second;
                        if(c>=a_num) break;
                        field[x][y] = a;
                        c++;
                        for(int l=0; l<4; ++l) {
                            int nx = x + dx[l];
                            int ny = y + dy[l];
                            if(nx>=0&&nx<h&&ny>=0&&ny<w&&field[nx][ny]==0) {
                                que.push(make_pair(nx, ny));
                            }
                        }
                    }
                    f++;
                    break;
                }
            }
        }
    }

    for(int i=0; i<h; ++i) {
        for(int j=0; j<w; ++j) {
            if(j==0) cout << field[i][j];
            else cout << " " << field[i][j];
        }
        cout << endl;
    }
    return 0;
}