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)

/// kAAzuki Hoshino
/*
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define int long long
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int,int> pi;
typedef vector<pi> vpi;
#define inf 1000000000
#define mod 1000000007
#define rep(i,a,b) for (int i = a; i <= b; i++)
#define rep1(i, a, b) for(int i =a; i >=b; i--)
#define dmp(x) cerr<<"line "<<__LINE__<<" "<<#x<<":"<<x<<endl
#define fs first
#define sc second
#define pb push_back
#define all(c) (c).begin(),(c).end()
#define mt make_tuple
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){
	return o<<"("<<p.fs<<","<<p.sc<<")";
}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){
	o<<"{";
	for(const T& v:vc) o<<v<<",";
	o<<"}";
	return o;
}

int h,w,n,ans[10];
vector<pair<int,int>> points,black;

signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);

  cin >> h >> w >> n;
  for(int i=1;i<=n;i++){
    int x,y;
    cin >> x >> y;
    black.pb({x,y});
    for(int dx=-2;dx<=0;dx++){
      for(int dy=-2;dy<=0;dy++){
        if( x+dx>=1 && x+dx+2<=h && y+dy>=1 && y+dy+2<=w   )
        points.pb({ x+dx, y+dy });
      }
    }
  }
  sort(all(points));
  sort(all(black));
  points.erase(unique(all(points)), points.end());

  for(auto t:points){
    int x = t.fs;
    int y = t.sc;
    int cnt = 0;
    for(int dx=0;dx<=2;dx++){
      for(int dy=0;dy<=2;dy++){
        if(binary_search( all(black) , make_pair( x+dx, y+dy ) ))cnt++;
      }
    }
    ans[cnt]++;
  }

  ans[0] = (h-2)*(w-2) - points.size();

  for(int i=0;i<=9;i++){
    cout << ans[i] << endl;
  }


return 0;
}
///....