#include <bits/stdc++.h>
using namespace std;
template <class T>
istream &operator>>(istream &in, pair<T, T> &p) {
  in >> p.first >> p.second;
  return in;
}
template <class T>
ostream &operator<<(ostream &out, pair<T, T> &p) {
  out << p.first << ' ' << p.second;
  return out;
}
template <class Tuple, std::size_t N>
struct TuplePrinter {
  static void print(ostream &out, const Tuple &t) {
    TuplePrinter<Tuple, N - 1>::print(out, t);
    out << ' ' << get<N - 1>(t);
  }
};
template <class Tuple>
struct TuplePrinter<Tuple, 1> {
  static void print(ostream &out, const Tuple &t) { out << get<0>(t); }
};
template <class... Args>
ostream &operator<<(ostream &out, const tuple<Args...> &p) {
  TuplePrinter<decltype(p), sizeof...(Args)>::print(out, p);
  return out;
}
long long N, M, K;
long long active[1005][1005];
long long dp[1005][1005][5];
long long ans = 0;
void setVal(long long r, long long c, long long last, long long val) {
  ans -= dp[r][c][last];
  dp[r][c][last] = val;
  ans += dp[r][c][last];
}
void prepare() {
  for (long long i = 1; i <= N; i++) {
    for (long long j = 1; j <= M; j++) {
      active[i][j] = 1;
      setVal(i, j, 0, dp[i][j - 1][1] + 1);
      setVal(i, j, 1, dp[i - 1][j][0] + 1);
    }
  }
  ans -= N * M;
}
void disable(long long r, long long c, long long last) {
  setVal(r, c, last, 0);
  active[r][c] = 0;
  while (r <= N && c <= M) {
    if (last == 0)
      r++;
    else
      c++;
    last ^= 1;
    if (active[r][c]) {
      if (last == 0)
        setVal(r, c, last, dp[r][c - 1][!last] + 1);
      else
        setVal(r, c, last, dp[r - 1][c][!last] + 1);
    } else
      break;
  }
}
void enable(long long r, long long c, long long last) {
  setVal(r, c, 0, dp[r][c - 1][1] + 1);
  setVal(r, c, 1, dp[r - 1][c][0] + 1);
  active[r][c] = 1;
  while (r <= N && c <= M) {
    if (last == 0)
      r++;
    else
      c++;
    last ^= 1;
    if (active[r][c])
      if (last == 0)
        setVal(r, c, last, dp[r][c - 1][!last] + 1);
      else
        setVal(r, c, last, dp[r - 1][c][!last] + 1);
    else
      break;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin >> N >> M >> K;
  prepare();
  long long r, c;
  while (K--) {
    cin >> r >> c;
    if (active[r][c]) {
      disable(r, c, 1);
      disable(r, c, 0);
      ans++;
      cout << ans << "\n";
    } else {
      enable(r, c, 1);
      enable(r, c, 0);
      ans--;
      cout << ans << "\n";
    }
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

You are given a matrix, consisting of n rows and m columns. The rows are numbered top to bottom, the columns are numbered left to right.

Each cell of the matrix can be either free or locked.

Let's call a path in the matrix a staircase if it: 

  * starts and ends in the free cell; 
  * visits only free cells; 
  * has one of the two following structures: 
    1. the second cell is 1 to the right from the first one, the third cell is 1 to the bottom from the second one, the fourth cell is 1 to the right from the third one, and so on; 
    2. the second cell is 1 to the bottom from the first one, the third cell is 1 to the right from the second one, the fourth cell is 1 to the bottom from the third one, and so on. 



In particular, a path, consisting of a single cell, is considered to be a staircase.

Here are some examples of staircases:

<image>

Initially all the cells of the matrix are free.

You have to process q queries, each of them flips the state of a single cell. So, if a cell is currently free, it makes it locked, and if a cell is currently locked, it makes it free.

Print the number of different staircases after each query. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.

Input

The first line contains three integers n, m and q (1 ≤ n, m ≤ 1000; 1 ≤ q ≤ 10^4) — the sizes of the matrix and the number of queries.

Each of the next q lines contains two integers x and y (1 ≤ x ≤ n; 1 ≤ y ≤ m) — the description of each query.

Output

Print q integers — the i-th value should be equal to the number of different staircases after i queries. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.

Examples

Input


2 2 8
1 1
1 1
1 1
2 2
1 1
1 2
2 1
1 1


Output


5
10
5
2
5
3
1
0


Input


3 4 10
1 4
1 2
2 3
1 2
2 3
3 2
1 3
3 4
1 3
3 1


Output


49
35
24
29
49
39
31
23
29
27


Input


1000 1000 2
239 634
239 634


Output


1332632508
1333333000

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
istream &operator>>(istream &in, pair<T, T> &p) {
  in >> p.first >> p.second;
  return in;
}
template <class T>
ostream &operator<<(ostream &out, pair<T, T> &p) {
  out << p.first << ' ' << p.second;
  return out;
}
template <class Tuple, std::size_t N>
struct TuplePrinter {
  static void print(ostream &out, const Tuple &t) {
    TuplePrinter<Tuple, N - 1>::print(out, t);
    out << ' ' << get<N - 1>(t);
  }
};
template <class Tuple>
struct TuplePrinter<Tuple, 1> {
  static void print(ostream &out, const Tuple &t) { out << get<0>(t); }
};
template <class... Args>
ostream &operator<<(ostream &out, const tuple<Args...> &p) {
  TuplePrinter<decltype(p), sizeof...(Args)>::print(out, p);
  return out;
}
long long N, M, K;
long long active[1005][1005];
long long dp[1005][1005][5];
long long ans = 0;
void setVal(long long r, long long c, long long last, long long val) {
  ans -= dp[r][c][last];
  dp[r][c][last] = val;
  ans += dp[r][c][last];
}
void prepare() {
  for (long long i = 1; i <= N; i++) {
    for (long long j = 1; j <= M; j++) {
      active[i][j] = 1;
      setVal(i, j, 0, dp[i][j - 1][1] + 1);
      setVal(i, j, 1, dp[i - 1][j][0] + 1);
    }
  }
  ans -= N * M;
}
void disable(long long r, long long c, long long last) {
  setVal(r, c, last, 0);
  active[r][c] = 0;
  while (r <= N && c <= M) {
    if (last == 0)
      r++;
    else
      c++;
    last ^= 1;
    if (active[r][c]) {
      if (last == 0)
        setVal(r, c, last, dp[r][c - 1][!last] + 1);
      else
        setVal(r, c, last, dp[r - 1][c][!last] + 1);
    } else
      break;
  }
}
void enable(long long r, long long c, long long last) {
  setVal(r, c, 0, dp[r][c - 1][1] + 1);
  setVal(r, c, 1, dp[r - 1][c][0] + 1);
  active[r][c] = 1;
  while (r <= N && c <= M) {
    if (last == 0)
      r++;
    else
      c++;
    last ^= 1;
    if (active[r][c])
      if (last == 0)
        setVal(r, c, last, dp[r][c - 1][!last] + 1);
      else
        setVal(r, c, last, dp[r - 1][c][!last] + 1);
    else
      break;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin >> N >> M >> K;
  prepare();
  long long r, c;
  while (K--) {
    cin >> r >> c;
    if (active[r][c]) {
      disable(r, c, 1);
      disable(r, c, 0);
      ans++;
      cout << ans << "\n";
    } else {
      enable(r, c, 1);
      enable(r, c, 0);
      ans--;
      cout << ans << "\n";
    }
  }
}

```