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: Time Limit Exceeded)

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

static const double EPS = 1e-8;
static const double PI = 4.0 * atan(1.0);
static const double PI2 = 8.0 * atan(1.0);

#define REP(i,n)	for(int i=0;i<(int)n;++i)
#define ALL(c)		(c).begin(),(c).end()
#define CLEAR(v)	memset(v,0,sizeof(v))
#define MP(a,b)		make_pair((a),(b))
#define ABS(a)		((a)>0?(a):-(a))
#define FOR(i,s,n)	for(int i=s;i<(int)n;++i)

typedef pair<int, int> P;

int H, W, N;
P ab[100000];
set<P> vted;
bool img[5][5];
ll res[10];

int main(int argc, char **argv) {
	cin >> H >> W >> N;
	REP(i, N) {
		cin >> ab[i].first >> ab[i].second;
		--ab[i].first; --ab[i].second;
	}
	sort(ab, ab + N);
	cerr << "[sort]" << endl;
	REP(i, N) cerr << ab[i].first << " " << ab[i].second << endl;
	CLEAR(res);
	REP(i, N) {
		CLEAR(img);
		P &p = ab[i];
		REP(a, 5) {
			auto top = lower_bound(ab, ab + N, P(p.first - 2 + a, p.second), [&](const P &a, const P &b) { return a.first < b.first; });
			auto btm = upper_bound(ab, ab + N, P(p.first - 2 + a, p.second), [&](const P &a, const P &b) { return a.first < b.first; });
			REP(b, 5) {
				if (binary_search(top, btm, P(p.first - 2 + a, p.second - 2 + b), [&](const P &a, const P &b) { return a.second < b.second; })) img[a][b] = true;
			}
		}
		cerr << "[img at (" << p.first << ", " << p.second << ")]" << endl;
		REP(a, 5) {
			REP(b, 5) cerr << (img[a][b] ? 'x' : '.');
			cerr << endl;
		}
		REP(a, 3) REP(b, 3) {
			int c = p.first - 2 + a, d = p.second - 2 + b;
			if (c < 0 || d < 0 || c >= H - 2 || d >= W - 2 || vted.find(P(c, d)) != vted.end()) continue;
			vted.insert(P(c, d));
			int cnt = 0;
			REP(i, 3) REP(j, 3) if (img[a + i][b + j]) ++cnt;
			++res[cnt];
			cerr << "[cnt at (" << c << ", " << d << ") = " << cnt << "]" << endl;
		}
	}
	cerr << "[result]" << endl;
	res[0] = ll(W - 2) * (H - 2);
	REP(i, 9) res[0] -= res[i + 1];
	REP(i, 10) cout << res[i] << endl;
	return 0;
}