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: Runtime Error)

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <vector>
using namespace std;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}

typedef long long ll;
ll MOD = 1000000007;

// a^p
// 2^3 = 2 * 2^2
// 2^2 = 2 * (2^1)
// 2^1 = 2
ll modpow(ll a, ll p, ll mod) {
  if (p == 0) return 1;

  if (p % 2 == 0) {
    ll half = modpow(a, p / 2, mod) % mod;
    return half * half % mod;
  } else {
    return a * modpow(a, p - 1, mod) % mod;
  }
}

// nCa を求める
ll modCombination(ll n, ll a, ll mod) {
  if (n - a < a) {
    return modCombination(n, n - a, mod);
  }

  ll denominator = 1;  // 分母
  ll numerator = 1;    // 分子

  for (ll i = 0; i < a; i++) {
    denominator *= a - i;
    numerator *= n - i;
    denominator %= mod;
    numerator %= mod;
  }

  return numerator * modpow(denominator, mod - 2, mod) % mod;
}

class UnionFind {
 private:
  vector<ll> parents;

 public:
  UnionFind(ll n) : parents(n, -1) {}

  bool issame(ll x, ll y) { return root(x) == root(y); }

  bool merge(ll x, ll y) {
    if (issame(x, y)) return false;

    ll rx = root(x);
    ll ry = root(y);
    if (parents[rx] > parents[ry]) swap(rx, ry);
    // サイズ情報を更新
    parents[rx] += parents[ry];
    // yの親を更新
    parents[ry] = rx;

    return true;
  }

  ll size(ll x) { return -parents[root(x)]; }

 private:
  ll root(ll x) {
    if (parents[x] < 0) return x;
    // 根の親の値に木の(-)サイズの情報を入れる
    return parents[x] = root(parents[x]);
  }
};

int main() {
  cout << fixed << setprecision(15);

  ll H, W;
  cin >> H >> W;

  ll N;
  cin >> N;

  vector<ll> A(N);
  for (ll i = 0; i < N; i++) {
    cin >> A[i];
  }

  vector<ll> coloring(H * W);
  ll index = 0;
  for (ll i = 0; i < N; i++) {
    ll a = A[i];
    for (ll j = 0; j < a; j++) {
      coloring[index++] = i;
    }
  }
  for (ll i = 0; i < H; i++) {
    for (ll j = 0; j < W; j++) {
      ll index = i * H + j;
      if (i % 2 != 0) {
        index = (i + 1) * W - index;
      }
      if (j == W - 1) {
        cout << coloring[index] + 1 << endl;
      } else {
        cout << coloring[index] + 1 << " ";
      }
    }
  }

  return 0;
}