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: Wrong Answer)

#include <bits/stdc++.h>
#include <string>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define POW(n) ((n)*(n))
#define ALL(a) (a).begin(),(a).end()
#define cerr if(false) cerr

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<unsigned long long> vull;

void solve(long long H, long long W, long long N, vector<long long> a, vector<long long> b){
	ll ans[10] = {};
	vector<ll> list[H];
	
	REP(i,N)
		list[a[i]].push_back(b[i]);
	
	REP(i,N)
		sort(ALL(list[i]));
	
	REP(y,H){
		int ny = y+2;
		if(ny >= H)continue;
		
		REP(x,W){
				int nx = x+2;
				if(nx >= x)continue;
				
				int count = 0;
				FOR(nny,y,ny+1){
					for(const ll& nnx:list[y]){
						if(nnx < x)continue;
						if(nnx > nx)break;
						count++;
					}
				}
				
				assert(count < 10);
				
				ans[count]+= count;
		}
	}
	
	REP(i,10)
		cout << ans[i] << endl;
		
	// REP(i,10)
		// cout << "aa" << endl;
}



int main(){	
	ios::sync_with_stdio(false);
	long long H;
	long long N;
	long long W;
	cin >> H;
	cin >> W;
	cin >> N;
	vector<long long> b(N-1+1);
	vector<long long> a(N-1+1);
	for(int i = 0 ; i <= N-1 ; i++){
		cin >> a[i];
		cin >> b[i];
	}
	solve(H, W, N, a, b);
	return 0;
}