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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

ll H,W,N;
unordered_map<ll,ll> um;

ll idx(ll h,ll w){
    return (h+1)*1e9 + w;
}

int main(){
    cin >> H >> W >> N;
    for(int i=0;i<N;i++){
        ll a,b;
        cin >> a >> b;
        for(int j=0;j<3;j++){
            for(int k=0;k<3;k++){
                ll aa = a - j;
                ll bb = b - k;
                if(aa >= 1 && bb >= 1 && aa <= H-2 && bb <= W-2){
                    um[idx(aa,bb)]++;
                }
            }
        }
    }
    vector<ll> ans(9,0);
    ll total = 0;
    for(auto &e : um){
        ans[e.second-1]++;
        total++;
    }
    cout << (H-2) * (W-2) - total << endl;
    for(auto &e : ans) cout << e << endl;
    return 0;
}