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;
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define rep(i, n) for (ll i = 0; i < n; i++)
#define Rep(i, r, n) for (ll i = r; i < n; i++)
#define pb push_back
#define MOD 1000000007
#define INF 1000000000000
#define EPS 0.00000000001
typedef long long ll;

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
    //---------------------------------------------
    int h,w;
    cin>>h>>w;
    int n;
    cin>>n;
    vector<pair<int,int>> a(n);
    rep(i,n){
        cin>>a[i].first;
        a[i].second=i+1;
    }
    sort(all(a));
    vector<int> x(w);
    int c=0;
    int memo=0;
    rep(i,n){
        rep(j,a[i].first){
            x[c]=a[i].second;
            c++;
            if(c==w){
                c=0;
                if(memo%2==1){
                    reverse(all(x));
                }
                rep(k,w){
                    cout<<x[k]<<" ";
                }
                cout<<endl;
                memo++;
            }
        }
    }
    

    
}