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)

#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>

#if __cplusplus >= 201103L
#include <ccomplex>
#include <cfenv>
#include <cinttypes>
#include <cstdbool>
#include <cstdint>
#include <ctgmath>
#include <cwchar>
#include <cwctype>
#endif

// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>

#if __cplusplus >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#endif
using namespace std;

#define ll long long
//typedef pair<int,int> Pint;
typedef pair<ll, ll> P;
//typedef pair<int, pair<int, int>> P;
//typedef tuple<int,int,int> T;
typedef vector<ll> vec;
typedef vector<vec> mat;
#define rep(i, n) for(ll i = 0; i < n; i++)
#define revrep(i, n) for(ll i = n-1; i >= 0; i--)
ll max(ll a, ll b){return (a > b) ? a : b;}
ll min(ll a, ll b){return (a < b) ? a : b;}
ll Mypow(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2) res *= x;
    x *= x;
    k /= 2;
  }
  return res;
}


ll INFL = 1LL << 60;//10^18 = 2^60
int INF = 1 << 30;//10^9
ll MOD  = 1000000007;
//vector<int> dy = {0,0,1,-1};
//vector<int> dx = {1,-1,0,0};

ll H, W, N;

int main(){
  cin >> H >> W >> N;
  vector<ll> A(N), B(N);
  set<pair<int, int>> ab;
  rep(i, N){
    cin >> A[i] >> B[i];
    A[i]--, B[i]--;
    ab.insert({A[i], B[i]});
  }

  set<pair<ll, ll>> s;//正方形の左上を保存
  rep(i, N){
    ll a = A[i]-2, b = B[i]-2;
    rep(h, 3){
      rep(w, 3){
        if(0 <= a+h && a+h+2 < H && 0 <= b+w && b+w+2 < W) s.insert({a+h, b+w});
      }
    }
  }
  vector<ll> ans(10, 0);
  for(auto itr = s.begin(); itr != s.end(); ++itr){
    pair<ll, ll> z = *itr;
    ll y = z.first, x = z.second;
    ll cnt = 0;
    rep(h, 3){
      rep(w, 3){
        if(ab.count({y+h, x+w})) cnt++;
      }
    }
    ans[cnt]++;
  }
  ll sum = 0;
  rep(i, 9) sum += ans[i];
  ans[0] = (H-2) * (W-2) - sum;
  rep(i, 10) cout << ans[i] << endl;


}