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)

//#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<ll,ll> pll;
#define mp make_pair
#define f first
#define s second
#define pb push_back

const int N = 1e4 + 5;
const int OO = 2e9;

int h,w,n,a[N],ans[101][101];

void init() {



}

int main() {

    ios_base::sync_with_stdio(false);
    cin.tie(0);

    cin >> h >> w >> n;

    for(int i=0;i<n;++i) {

        cin >> a[i];

    }

    int cur = 0;

    for(int i=0;i<h;++i) {

        if(i&1) {

            for (int j = 0; j < w; ++j) {

                if (!a[cur]) ++cur;
                ans[i][j] = cur + 1;
                a[cur]--;

            }

        } else {

            for (int j = w-1; j >= 0; --j) {

                if (!a[cur]) ++cur;
                ans[i][j] = cur + 1;
                a[cur]--;

            }

        }

    }

    for(int i=0;i<h;++i) {

        for(int j=0;j<w;++j) {

            cout << ans[i][j] << ' ';

        }

        cout << '\n';

    }

    return 0;

}