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>
#define FOR(i,a,b) for(LL i=(a);i<(LL)(b);i++)
#define REP(i,b) FOR(i,0,b)
#define ALL(c) c.begin(),c.end()
#define PB push_back
#define FF first
#define SS second
#define cat //cout << __LINE__ << endl;
using namespace std;
typedef long long LL;
typedef double ld;
typedef LL ut;
typedef vector<ut> VI;
typedef pair<ut,ut> pr;
typedef pair<ut,pr> ppr;
typedef vector<pr> Vpr;
typedef priority_queue<pr,Vpr> PQ;
const ut INF=1LL<<60;
const LL p=3+1e6;
const int SIZE=105;
map<pr,LL> maps;
int main(){
	LL H,W,N,a,b;
	cin >> H >> W >> N;
	LL ans[10]={};
	ans[0]=(H-2)*(W-2);
	cin.tie(0);
	REP(i,N){
		cin >> a >> b;
		FOR(i,a-1,a+2) FOR(j,b-1,b+2){
			if(2<=i && i<=H-1 && 2<=j && j<=W-1)
				maps[pr(i,j)]=maps[pr(i,j)]+1;
		}
	}
	for(auto it=maps.begin();it!=maps.end();++it){
		ans[(*it).second]++;
		ans[0]--;
	}
	REP(i,10) cout << ans[i] << endl;
	return 0;
}