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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <cassert>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <iostream>
#include <vector>
#include <string>
#include <sstream>
#include <algorithm>
#include <utility>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <functional>
#include <numeric>
#include <complex>
#include <iomanip>

using namespace std;

#define bitcount(n) __builtin_popcount(n)

#define debug(n) cout << n << endl;

#define PREC(n) setprecision(n)
#define DOTPREC(n) fixed << setprecision(n)

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

#define fst first
#define scd second
#define PB push_back
#define MP make_pair
#define rep(i,x) for(int i=0;i<(x);++i)
#define rep1(i,x) for(int i=1;i<=(x);++i)
#define rrep(i,x) for(int i=(x)-1;i>=0;--i)
#define rrep1(i,x) for(int i=(x);i>=1;--i)
#define FOR(i,a,x) for(int i=(a);i<(x);++i)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()

#define omajinai ios::sync_with_stdio(false); cin.tie(0)

template<typename T> bool chmax(T&a,T b) { if(a < b) {a = b; return true;} return false;}
template<typename T> bool chmin(T&a,T b) { if(a > b) {a = b; return true;} return false;}

const int inf = 1e9;
const ll linf = 3e18;
const double eps = 1e-9;

const ll MOD = 1000000007;

#define INDEX_MAX 100000
#define SIZE 1000000000

ll H, W, N;

typedef unordered_map<ll, unordered_map<ll, bool>> BOARD;
BOARD blacks;
BOARD will_check;

int count(int i, int j) {
    int total = 1;
    for (int k = -1; k <= 1; k++) {
        for (int l = -1; l <= 1; l++) {
            if (i + k < 0 || i + k >= H) return -1;
            if (j + l < 0 || j + l >= W) return -1;

            total +=
                blacks.find(i+k) != blacks.end()
                && blacks[i+k].find(j+l) != blacks[i+k].end();
        }
    }
    return total;
}

void mark(int i, int j) {
    for (int k = -1; k <= 1; k++) {
        for (int l = -1; l <= 1; l++) {
            if (i + k < 0 || i + k >= H) continue;
            if (j + l < 0 || j + l >= W) continue;
            will_check[i+k][j+l] = true;
        }
    }
}

int main(void)
{
    cin >> H >> W >> N;

    rep(i, N) {
        ll a, b;
        scanf("%lld %lld", &a, &b);
        a--, b--;
        blacks[a][b] = true;
        mark(a, b);
    }

    ll counts[10] = {};
    for (auto& i_map : will_check) {
        const int i = i_map.first;
        for (auto& j_map : i_map.second) {
            const int j = j_map.first;
            const int c = count(i, j);
            if (c != -1) counts[c]++;
        }
    }

    ll total = (H - 2) * (W - 2);
    for (auto& count : counts) {
        total -= count;
    }
    counts[0] = total;

    rep(i, 10) {
        cout << counts[i] << endl;
    }
    return 0;
}