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 <bits/stdc++.h>
using namespace std;

typedef pair<int, int> P;

#define ft first
#define sd second

int h, w;
int n;
long long res[10];
P p[100055];
set<P> s;
set<P> used;

void go(int x, int y) {
	if (x < 1 || x + 2 > h || y < 1 || y + 2 > w) return;
	if (used.find(P(x, y)) != used.end()) return;

	used.insert(P(x, y));

	int cnt = 0;

	for (int i = x; i <= x + 2; i++) {
		for (int j = y; j <= y + 2; j++) {
			if (s.find(P(i, j)) != s.end()) {
				cnt++;
			}
		}
	}

	res[cnt]++;
}
int main() {
	scanf("%d %d", &h, &w);
	scanf("%d", &n);

	for (int i = 0; i < n; i++) {
		scanf("%d %d", &p[i].ft, &p[i].sd);
		s.insert(p[i]);
	}

	for (int i = 0; i < n; i++) {
		for (int x = p[i].ft - 2; x <= p[i].ft; x++) {
			for (int y = p[i].sd - 2; y <= p[i].sd; y++) {
				go(x, y);
			}
		}
	}

	long long sum = 0;

	for (int i = 1; i < 10; i++) sum += res[i];

	res[0] = 1ll * (h - 2) * (w - 2) - sum;

	for (int i = 0; i < 10; i++) printf("%lld\n", res[i]);

	return 0;
}