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 (Clang 3.8.0) (judge verdict: Accepted)

/*
このコード、と～おれ!
Be accepted!
∧＿∧　
（｡･ω･｡)つ━☆・*。
⊂　　 ノ 　　　・゜+.
　しーＪ　　　°。+ *´¨)
 　　　　　　　　　.· ´¸.·*´¨) ¸.·*¨)
		  　　　　　　　　　　(¸.·´ (¸.·'* ☆
*/

#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <vector>
#include <numeric>
#include <iostream>
#include <random>
#include <map>
#include <unordered_map>
#include <queue>
#include <regex>
#include <functional>
#include <complex>
#include <list>
#include <cassert>
#include <iomanip>
#include <set>
/*多倍長整数/cpp_intで宣言
#include <boost/multiprecision/cpp_int.hpp>
using namespace boost::multiprecision;
*/

//#pragma gcc target ("avx2")
//#pragma gcc optimization ("o3")
//#pragma gcc optimization ("unroll-loops")
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define rep1(i, n) for(int i = 1; i <= (n); ++i)
#define rep2(i, n) for(int i = 2; i < (n); ++i)
#define repr(i, n) for(int i = n; i >= 0; --i)
#define reprm(i, n) for(int i = n - 1; i >= 0; --i)
#define printynl(a) printf(a ? "yes\n" : "no\n")
#define printyn(a) printf(a ? "Yes\n" : "No\n")
#define printYN(a) printf(a ? "YES\n" : "NO\n")
#define printim(a) printf(a ? "possible\n" : "imposible\n")
#define printdb(a) printf("%.50lf\n", a) //少数出力
#define printdbd(a) printf("%.16lf\n", a) //少数出力(桁少なめ)
#define prints(s) printf("%s\n", s.c_str()) //string出力
#define all(x) (x).begin(), (x).end()
#define allsum(a, b, c) ((a + b) * c / 2.0) //等差数列の和、初項,末項,項数
#define pb push_back
#define priq priority_queue
#define rpriq priq<int, vector<int>, greater<int>>
#define deg_to_rad(deg) (((deg)/360.0)*2.0*PI)
#define rad_to_deg(rad) (((rad)/2.0/PI)*360.0)
#define Please return
#define AC 0
#define addf(T) [](T a, T b){return (a + b);}
#define minf(T) [](T a, T b){return min(a, b);}
#define maxf(T) [](T a, T b){return max(a, b);}
#define nopf(T) [](T a){return a;}
#define manhattan_dist(a, b, c, d) (abs(a - c) + abs(b - d)) /*(a, b) から (c, d) のマンハッタン距離 */


using ll = long long;

constexpr int INF = 1073741823;
constexpr int MINF = -1073741823;
constexpr ll LINF = ll(4661686018427387903);
constexpr ll MOD = 1000000007;
const long double PI = acos(-1.0L);

using namespace std;

void scans(string& str) {
	char c;
	str = "";
	scanf("%c", &c);
	if (c == '\n')scanf("%c", &c);
	while (c != '\n' && c != -1 && c != ' ') {
		str += c;
		scanf("%c", &c);
	}
}

void scanc(char& str) {
	char c;
	scanf("%c", &c);
	if (c == -1)return;
	while (c == '\n') {
		scanf("%c", &c);
	}
	str = c;
}

double acot(double x) {
	return PI / 2 - atan(x);
}

ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}

ll lcm(ll number1, ll number2) {
	return number1 / gcd(number1, number2) * number2;
}

ll LSB(ll n) { return (n & (-n)); }

/*-----------------------------------------ここからコード-----------------------------------------*/

int main() {

	int h, w, n;
	scanf("%d%d%d", &h, &w, &n);
	vector<int> a(n);
	rep(i, n) {
		scanf("%d", &a[i]);
	}
	vector<int> ans(h * w);
	int idx = 0;
	rep(i, n) {
		fill(ans.begin() + idx, ans.begin() + idx + a[i], i + 1);
		idx += a[i];
	}
	rep(i, h) {
		rep(j, w) {
			if (i & 1) {
				printf("%d\n", ans[i * w + (w - j - 1)]);
			}
			else printf("%d\n", ans[i * w + j]);
		}
	}

	Please AC;
}