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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "debug.h"
#else
#define dump(...)
#endif
#define endl '\n'
#define ll long long
#define FOR(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define REP(i, x) for (int i = 0; i < (int)(x); i++)
#define REPS(i, x) for (int i = 1; i <= (int)(x); i++)
#define RREP(i, x) for (int i = ((int)(x)-1); i >= 0; i--)
#define RREPS(i, x) for (int i = ((int)(x)); i > 0; i--)
#define INF 2147483647
#define LLINF 9223372036854775807LL
#define vi vector<int>
#define vvi vector<vector<int>>
#define pi pair<int, int>
#define ALL(a) (a).begin(), (a).end()
#define BIT(n) (1LL << (n))
#define TOUPPER(s) transform(s.begin(), s.end(), s.begin(), ::toupper)
#define TOLOWER(s) transform(s.begin(), s.end(), s.begin(), ::tolower)
#define FILLZERO(s, n)                      \
    do {                                    \
        ostringstream os;                   \
        os << setw(n) << setfill('0') << s; \
        s = os.str();                       \
    } while (0);
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end()) //sortしてからつかうこと
#define INIT                               \
    std::cin.tie(nullptr);                 \
    std::ios_base::sync_with_stdio(false); \
    cout << std::fixed << std::setprecision(10);

template <typename T>
vector<T> make_vec(size_t a, T n) { return vector<T>(a, n); }
template <typename T, typename... Ts>
auto make_vec(size_t a, Ts... ts) { return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...)); }
template <class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
    for (T &x : vec)
        is >> x;
    return is;
}

constexpr ll MOD = 1e9 + 7;
int dx[9] = {0, 1, 0, -1, 0, 1, 1, -1, -1};
int dy[9] = {0, 0, 1, 0, -1, 1, -1, 1, -1};
ll A, B, C, D, E, F, G, H, N, M, L, K, P, Q, R, W, X, Y, Z;
string S, T;
ll ans = 0;

template <typename T>
T intceil(T a, T b) {
    return (a + (b - 1)) / b;
}

signed main() {
    INIT;
    cin >> H >> W >> N;
    vi a(N), b(N);
    map<pair<int, int>, int> mp;
    REP(i, N) {
        cin >> a[i] >> b[i];
        a[i]--, b[i]--;
        mp[{a[i], b[i]}] = 1;
    }
    vector<ll> v(10);
    set<pair<int, int>> st;
    REP(i, N) {
        REP(j, 9) {
            int h = a[i] + dx[j];
            int w = b[i] + dy[j];
            if (0 < h && h < H - 1 && 0 < w && w < W - 1 && st.count({h, w}) == 0) {
                st.insert({h, w});
                int tmp = 0;
                REP(k, 9)
                tmp += mp[{h + dx[k], w + dy[k]}];
                v[tmp]++;
            }
        }
    }
    v[0] = (H - 2) * (W - 2) - st.size();
    dump(v);
    REP(i, v.size())
        cout << v[i] << endl;
}