Problem:
Score : 400 points

Problem Statement

We have a grid with H rows and W columns. At first, all cells were painted white.

Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column.

Compute the following:

For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?

Constraints

3 \leq H \leq 10^9

3 \leq W \leq 10^9

0 \leq N \leq min(10^5,H×W)

1 \leq a_i \leq H (1 \leq i \leq N)

1 \leq b_i \leq W (1 \leq i \leq N)

(a_i, b_i) \neq (a_j, b_j) (i \neq j)

Input

The input is given from Standard Input in the following format:

H W N
a_1 b_1
:
a_N b_N

Output

Print 10 lines.
The (j+1)-th ( 0 \leq j \leq 9 ) line should contain the number of the subrectangles of size 3×3 of the grid that contains exactly j black cells.

Sample Input 1

4 5 8
1 1
1 4
1 5
2 3
3 1
3 2
3 4
4 4

Sample Output 1

0
0
0
2
4
0
0
0
0
0

There are six subrectangles of size 3×3. Two of them contain three black cells each, and the remaining four contain four black cells each.

Sample Input 2

10 10 20
1 1
1 4
1 9
2 5
3 10
4 2
4 7
5 9
6 4
6 6
6 7
7 1
7 3
7 7
8 1
8 5
8 10
9 2
10 4
10 9

Sample Output 2

4
26
22
10
2
0
0
0
0
0

Sample Input 3

1000000000 1000000000 0

Sample Output 3

999999996000000004
0
0
0
0
0
0
0
0
0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cmath>
#include <set>
#include <queue>
#include <deque>
#include <vector>
#include <climits>
#include <sstream>
#include <iomanip>
#include <map>
#include <stack>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;

int main()
{
	ll H, W;
	int N;
	cin >> H >> W >> N;

	map<ll, vector<ll>> m;
	for (auto i = 0; i < N; ++i)
	{
		ll a, b;
		cin >> a >> b;
		--a;
		--b;
		m[b].push_back(a);
	}

	for (auto& p : m)
	{
		sort(p.second.begin(), p.second.end());
	}

	vector<ll> ans(10);

	for (auto& p : m)
	{
		auto x = p.first;
		for (auto y : p.second)
		{
			for (auto i = -2; i <= 0; ++i)
			{
				for (auto j = -2; j <= 0; ++j)
				{
					auto baseX = x + i;
					auto baseY = y + j;
					if (baseX < 0 || baseY < 0 || baseX + 2 >= W || baseY + 2 >= H)
					{
						continue;
					}

					auto valid = true;
					auto num = 1;
					for (auto k = 0; k < 3; ++k)
					{
						for (auto l = 0; l < 3; ++l)
						{
							auto tempX = baseX + k;
							auto tempY = baseY + l;
							if (tempX == x && tempY == y)
							{
								continue;
							}

							auto it = m.find(tempX);
							if (it == m.end())
							{
								continue;
							}

							auto it2 = lower_bound(it->second.begin(), it->second.end(), tempY);
							if (it2 == it->second.end()
								|| *it2 != tempY)
							{
								continue;
							}

							++num;
							if (tempX < x
								|| (tempX == x && tempY < y))
							{
								valid = false;
								break;
							}
						}
						if (!valid)
						{
							break;
						}
					}

					if (valid)
					{
						++ans[num];
					}
				}
			}
		}
	}

	auto sum = 0ll;
	for (auto i = 1; i < 10; ++i)
	{
		sum += ans[i];
	}
	ans[0] = (H - 2) * (W - 2) - sum;

	for (auto i = 0; i < 10; ++i)
	{
		cout << ans[i] << endl;
	}

	return 0;
}