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>


#define rep2(x,fr,to) for(int (x)=(fr);(x)<(to);(x)++)
#define rep(x,to) for(int (x)=0;(x)<(to);(x)++)
#define repr(x,fr,to) for(int (x)=(fr);(x)>=(to);(x)--)
#define all(c) (c).begin(),(c).end()
#define sz(v) (int)(v).size()

using namespace std;
typedef long long ll; typedef vector<int> VI; typedef pair<int,int> pii;
const int MD = (int)1e9 + 7; typedef vector<ll> VL;
void dbg(){ cerr << "\n"; }  template <class T, class ...T2>
void dbg(const T& fst, const T2&...rst){ cerr << fst << ": "; dbg(rst...); }
template <class T, class T2> void amax(T& a,T2 b){ if(a < b) a = b;}

int main()
{
	cin.tie(0); ios_base::sync_with_stdio(false);
	int h, w, n;
	cin >>h >>w >>n;
	VI a(n);
	rep(i, n) cin >>a[i];
	vector<VI> ba(h, VI(w));
	int cr = 0;
	rep(i, n){
		rep(j, a[i]){
			int y=cr/w, x = y%2?  w-1 -cr%w: cr%w;
			ba[y][x] = i+1;
			cr++;
		}
	}
	for(auto xl: ba){for(auto x :xl) cout<<x<<" "; cout<<"\n";}

	return 0;
}