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<iostream>
#include<queue>
using namespace std;

int main(){
    int h,w,n;cin >> h >> w >> n;
    queue<int> q;
    int ans[h][w];
    for(int i=0;i<n;i++){
        int c;
        cin >> c;
        for(int j=0;j<c;j++){
            q.push(i+1);
        }   
    }
    for(int i=0;i<h;i++){
        for(int j=0;j<w;j++){
            int now = q.front();
            #if TEST
                cout << now << endl;
            #endif
            q.pop();
            if(i%2 == 0){
                ans[i][j] =  now;
            }    
            else{
                ans[i][w-1-j] = now;
            }
        }
    }
    for(int i=0;i<h;i++){
        for(int j=0;j<w;j++){
            cout << ans[i][j];
            if(j == w-1){
                cout <<endl;
            }
            else{
                cout << " " ;
            }
        }
    }
    return 0;
}