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)

#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#define ll long long
#define rep2(i,a,b) for(int i=a;i<=b;++i)
#define rep(i,n) for(int i=0;i<n;i++)
#define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;

const int N=1e6+10;
ll a[N];
int sum[11];
main(){
    ios
    int h,w,n;cin>>h>>w>>n;
    h-=2,w-=2;
    int x,y;
    int cnt=0;
    ll m=(ll)h*w;
    rep(iii,n){
        cin>>x>>y;
        rep(i,3)if( x-i>=1 && x-i<=h){
            rep(j,3)if(y-j>=1 && y-j<=w){
            a[cnt++]=((ll)(x-i)<<32)+(y-j);
            }
        }
    }
    sort(a,a+cnt);
    ll t=1;
    rep(i,cnt){
        if(a[i]==a[i+1])t++;
        else sum[t]++,t=1,m--;
    }
    cout<<m<<" ";
    rep2(i,1,9)cout<<sum[i]<<" ";
}