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: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long LL;

LL ans[15], d[10000000], tot = 0;

LL chuli(int a, int b)
{
    LL mod = 20000222;
    a *= mod;
    b *= 521;
    return a + b;
}

int main()
{
    LL h, w, n, a, b;
    cin >> h >> w >> n;
    LL sy = (h-2)*(w-2);
    while (n--)
    {
        scanf("%lld%lld", &a, &b);
        for (LL i = 0; i < 3; i++)
        {
            for (LL j = 0; j < 3; j++)
            {
                if (a - i >= 1 && a - i <= h - 2 && b - j >= 1 && b - j <= w - 2)
                {
                    d[++tot] = chuli((a-i), (b-j));
                }
            }
        }
    }
    sort(d+1, d+tot+1);
    int ls = 1;
    for (int i = 1; i <= tot; i++)
    {
        if (d[i] == d[i+1])
        {
            ls++;
        }
        else
        {
            ans[ls]++;
            ls = 1;
            sy--;
        }
    }
    cout << sy << endl;
    for (int i = 1; i <= 9; i++)
    {
        cout << ans[i] << endl;
    }
    return 0;
}