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: Accepted)


#pragma GCC optimize("O3") //コンパイラ最適化用

#define _GLIBCXX_DEBUG //配列に[]でアクセス時のエラー表示
#include <algorithm>   //sort,二分探索,など
#include <bitset>      //固定長bit集合
#include <cassert>     //assert(p)で，not pのときにエラー
#include <chrono>      //実行時間計測
#include <cmath>       //pow,logなど
#include <complex>     //複素数
#include <cstring>
#include <deque>
#include <functional> //sortのgreater
#include <iomanip>    //setprecision(浮動小数点の出力の誤差)
#include <iostream>   //入出力
#include <iterator>   //集合演算(積集合,和集合,差集合など)
#include <map>
#include <numeric> //iota(整数列の生成),gcdとlcm,accumulate
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility> //pair
#include <vector>  //可変長配列
using namespace std;
typedef long long ll;
typedef long double ld;

#define ALL(x) x.begin(), x.end()
#define SIZE(x) ll(x.size()) // sizeをsize_tからllに直しておく
const long long INF = 1e12;
const int MOD = 1e9 + 7;
#define MAXR 100000 // 10^5:配列の最大のrange(素数列挙などで使用)
//略記
#define Umap unordered_map
#define Uset unordered_set
#define int long long
#define LD long double
#define P pair<int, int>
#define PB push_back
#define MP make_pair
#define VI vector<int>
#define VVI vector<vector<int>>
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define REPD(i, n) for (int i = (int)(n)-1; i >= 0; i--)
#define FOR(i, a, b) for (int i = a; i < (int)(b); i++)
#define FORD(i, a, b) for (int i = (int)(b)-1; i >= (int)a; i--)

const unsigned int BIT_FLAG_0 = (1 << 0); // 0000 0000 0000 0001
const unsigned int BIT_FLAG_1 = (1 << 1); // 0000 0000 0000 0010
const unsigned int BIT_FLAG_2 = (1 << 2); // 0000 0000 0000 0100
const unsigned int BIT_FLAG_3 = (1 << 3); // 0000 0000 0000 1000
const unsigned int BIT_FLAG_4 = (1 << 4); // 0000 0000 0001 0000
const unsigned int BIT_FLAG_5 = (1 << 5); // 0000 0000 0010 0000
const unsigned int BIT_FLAG_6 = (1 << 6); // 0000 0000 0100 0000
const unsigned int BIT_FLAG_7 = (1 << 7); // 0000 0000 1000 0000
const int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const int Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1},
          Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};

// https://qiita.com/Lily0727K/items/06cb1d6da8a436369eed#c%E3%81%A7%E3%81%AE%E5%AE%9F%E8%A3%85
void print() { cout << endl; }

template <class Head, class... Tail> void print(Head &&head, Tail &&... tail) {
    cout << head;
    if (sizeof...(tail) != 0)
        cout << " ";
    print(forward<Tail>(tail)...);
}

template <class T> void print(vector<T> &vec) {
    for (auto &a : vec) {
        cout << a;
        if (&a != &vec.back())
            cout << " ";
    }
    cout << endl;
}

template <class T> void print(vector<vector<T>> &df) {
    for (auto &vec : df) {
        print(vec);
    }
}
long long power(long long a, long long n) {
    long long res = 1;
    while (n > 0) {
        if (n & 1)
            res = res * a;
        a *= a;
        n >>= 1;
    }
    return res;
}
long long comb(int n, int k) {
    vector<vector<long long>> v(n + 1, vector<long long>(n + 1, 0));

    for (int i = 0; i < v.size(); i++) {
        v[i][0] = 1;
        v[i][i] = 1;
    }
    for (int k = 1; k < v.size(); k++) {
        for (int j = 1; j < k; j++) {
            v[k][j] = (v[k - 1][j - 1] + v[k - 1][j]);
        }
    }
    return v[n][k];
}
void add(long long &a, long long b) {
    a += b;
    if (a >= MOD)
        a -= MOD;
}
template <class T> inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

int h, w, n, line = 0, cur = 0;
VI A;
VVI res;
signed main() {
    cin >> h >> w >> n;
    FOR(i, 1, n + 1) {
        int times;
        cin >> times;
        REP(t, times) { A.PB(i); }
    }
    res.resize(h, vector<int>(w));

    REP(i, h * w) {
        if (line % 2 == 0) {
            if (cur < w - 1) {
                res[line][cur] = A[i];
                cur++;
            } else {
                res[line][cur] = A[i];

                line++;
            }
        }

        else {
            if (cur > 0) {
                res[line][cur] = A[i];
                cur--;
            } else {
                res[line][cur] = A[i];

                line++;
            }
        }
    }

    REP(hi, h) {
        REP(wi, w) {
            cout << res[hi][wi] << " ";
            if (wi == w - 1) {
                cout << endl;
            }
        }
    }
}