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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
int H, W;
int mass[110][110];

int dh[4] = {1, 0, -1, 0};
int dw[4] = {0, 1, 0, -1};

//色 n をまだ塗っていないところに p 個ぬる
void BFS(int n, int p)  {
  
  //どこから塗るか，スタートを決める
  pair<int, int> start;
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      if (mass[i][j] == 0) {
        start = make_pair(i,j);
        break;
      }
    }
  }
  
  queue<pair<int,int>> que;
  que.push(start);
  
  mass[start.first][start.second] = n;
  int count = 1;
  while(que.size()) {
    pair<int, int> v = que.front();
    que.pop();
    
    pair<int, int> nv;
    for (int i = 0; i < 4 ; i++) {
      nv = make_pair(v.first + dh[i], v.second + dw[i]);
      int nh = nv.first, nw = nv.second;
      
      if (nh >= 0 && nh < H && nw >= 0 && nw < W && mass[nh][nw] == 0) {
        que.push(nv);
        mass[nh][nw] = n;
        count++;
      }
      
      if (count == p) {
        break;
      }
      
    }
    
    if (count == p) {
       break;
    }
  }
  
       
       
}

int main() {
  cin >> H >> W;
  
  int N;
  cin >> N;
  
  vector<int> a(N+1);
  for (int i = 1; i <= N; i++) {
    cin >> a[i];
  }
  
  
  
  //まだ塗っていないところは0
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      mass[i][j] = 0;
    }
  }
  
  
  
  // BFSで色ぬりする
  for (int i = 1; i <= N; i++) {

    //色 i を a[i] 個塗る
    BFS(i, a[i]);
  }
  
  // 結果の出力
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      cout << mass[i][j];
    }
    cout << endl; 
  }
}