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 <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp>*/
#include <bits/stdc++.h>

using namespace std;
//using namespace __gnu_pbds;
 
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;

#define FOR(i, a, b) for (int i=a; i<b; i++)
#define F0R(i, a) for (int i=0; i<a; i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
 
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound

const int MOD = 1000000007;
double PI = 4*atan(1);

int H,W,N, grid[100][100];
vector<pii> ok;

int main() {
	ios_base::sync_with_stdio(0);cin.tie(0);
	cin >> H >> W >> N;
	F0R(i,H) {
	    if (i % 2 == 0) F0R(j,W) ok.pb({i,j});
	    else F0Rd(j,W) ok.pb({i,j});
	}
	int lst = 0;
	FOR(i,1,N+1) {
	    int x; cin >> x;
	    FOR(j,lst,lst+x) grid[ok[j].f][ok[j].s] = i;
	    lst += x;
	}
	F0R(i,H) {
	    F0R(j,W) cout << grid[i][j] << " ";
	    cout << "\n";
	}
}