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)

/* Generated by powerful Codeforces Tool
 * You can download the binary file in here https://github.com/xalanq/cf-tool (Windows, macOS, Linux)
 * Author: alireza_kaviani
 * Time: 2020-06-08 13:52:16
**/
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;

#define all(x)                      (x).begin(),(x).end()
#define Sort(x)                     sort(all((x)))
#define X                           first
#define Y                           second
#define sep                         ' '
#define endl                        '\n'
#define SZ(x)                       ll(x.size())

ll poww(ll a, ll b, ll md) {
    return (!b ? 1 : (b & 1 ? a * poww(a * a % md, b / 2, md) % md : poww(a * a % md, b / 2, md) % md));
}

const ll MAXN = 1e6 + 10;
const ll LOG = 22;
const ll INF = 8e18;
const ll MOD = 1e9 + 7; // 998244353; // 1e9 + 9;

ll H , W , n , ans[10];
vector<pll> vec;
map<pll , ll> mark;

int main() {
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

    cin >> H >> W >> n;
    for(int i = 0 , x , y ; i < n ; i++){
        cin >> x >> y;
        for(int j = 0 ; j < 3 ; j++){
            for(int k = 0 ; k < 3 ; k++){
                int nx = x - j , ny = y - k;
                if(nx <= 0 || ny <= 0 || nx + 2 > H || ny + 2 > W)  continue;
                vec.push_back({nx , ny});
            }
        }
        mark[{x , y}] = 1;
    }
    sort(all(vec));
    vec.resize(unique(all(vec)) - vec.begin());
    for(pll i : vec){
        int x = i.X , y = i.Y , c = 0;
        //cout << x << sep << y << endl;
        for(int j = 0 ; j < 3 ; j++)    for(int k = 0 ; k < 3 ; k++)    c += mark[{x + j , y + k}];
        ans[c]++;
    }
    ans[0] += (H - 2) * (W - 2) - SZ(vec);
    for(int i = 0 ; i < 10 ; i++)   cout << ans[i] << endl;
    
    return 0;
}
/*

*/