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 <queue>
#include <map>
#include <list>
#include <deque>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
#define dump(a) cout << (a) << endl;
typedef long long ll;
ll mod = 1e9+7;

int main() {
    int h, w, n;
    cin >> h >> w >> n;
    vector<int> a(n+1, 0);
    for( int i = 1; i <= n; i++ ) cin >> a[i];
    for( int i = 0; i < n; i++ ) a[i+1] += a[i];
    vector<int> f( h*w );
    for( int i = 0; i < n; i++ ) for( int j = a[i]; j < a[i+1]; j++ ) f[j] = i+1;
    vector<vector<int>> g(h, vector<int>(w));
    int x = 0;
    int i, j;
    int k = 0;
    while( w-x*2 > 0 && h-x*2 > 0 ) {
        i = j = x;
        while( j < w-x ) {
            g[i][j] = f[k];
            k++;
            j++;
        }
        j--;
        k--;
        while( i < h-x-1 ) {
            g[i][j] = f[k];
            k++;
            i++;
        }
        if( h-x*2 == 1 ) break;
        while( x < j ) {
            g[i][j] = f[k];
            k++;
            j--;
        }
        if( w-x*2 == 1 ) break;
        while( x < i ) {
            g[i][j] = f[k];
            k++;
            i--;
        }   
        x++;
    }
    for( int i = 0; i < h; i++ ) {
        for( int j = 0; j < w; j++ ) {
            cout << g[i][j] << " ";
        }
        cout << endl;
    }
    return 0;  
}