#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
string dirs = "DRUL";
bool Free[N][N];
int dir[N][N][4];
int Freecnt = 0;
long long sum01 = 0;
int n, m, q;
bool isFree(int x, int y) {
  return (1 <= x && x <= n && 1 <= y && y <= m && Free[x][y]);
}
int cnt(int x, int y, int dir) {
  if (!isFree(x, y)) return 0;
  return 1 + cnt(x + dx[dir], y + dy[dir], dir ^ 1);
}
void add(int x, int y, int dir, int dz) {
  if (!isFree(x, y)) return;
  ::dir[x][y][dir ^ 2] += dz;
  if (dir >= 2) sum01 += dz;
  return add(x + dx[dir], y + dy[dir], dir ^ 1, dz);
}
void flip(int x, int y) {
  if (Free[x][y] == 0) {
    Free[x][y] = 1;
    Freecnt++;
    for (int dd = 0; dd < 4; dd++) {
      int cc = cnt(x, y, dd);
      add(x, y, dd ^ 3, cc);
    }
  } else {
    for (int dd = 0; dd < 4; dd++) {
      int cc = cnt(x, y, dd);
      add(x, y, dd ^ 3, -cc);
    }
    Free[x][y] = 0;
    Freecnt--;
  }
}
void init() {
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      Free[i][j] = 1;
      Freecnt++;
      dir[i][j][2] = dir[i][j][3] = 1;
      if (i) dir[i][j][2] += dir[i - 1][j][3];
      if (j) dir[i][j][3] += dir[i][j - 1][2];
    }
  }
  for (int i = n; i >= 1; i--) {
    for (int j = m; j >= 1; j--) {
      dir[i][j][0] = dir[i][j][1] = 1;
      if (i < n) dir[i][j][0] += dir[i + 1][j][1];
      if (j < m) dir[i][j][1] += dir[i][j + 1][0];
      sum01 += dir[i][j][0] + dir[i][j][1];
    }
  }
}
void all() {
  cout << sum01 << " " << Freecnt << " " << sum01 - Freecnt << endl;
  for (int d = 0; d < 4; d++) {
    cout << d << " -----> " << endl;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        cout << dir[i][j][d] << " ";
      }
      cout << endl;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  init();
  while (q--) {
    int x, y;
    cin >> x >> y;
    flip(x, y);
    cout << sum01 - Freecnt << "\n";
    cout.flush();
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int N = 1005;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
string dirs = "DRUL";
bool Free[N][N];
int dir[N][N][4];
int Freecnt = 0;
long long sum01 = 0;
int n, m, q;
bool isFree(int x, int y) {
  return (1 <= x && x <= n && 1 <= y && y <= m && Free[x][y]);
}
int cnt(int x, int y, int dir) {
  if (!isFree(x, y)) return 0;
  return 1 + cnt(x + dx[dir], y + dy[dir], dir ^ 1);
}
void add(int x, int y, int dir, int dz) {
  if (!isFree(x, y)) return;
  ::dir[x][y][dir ^ 2] += dz;
  if (dir >= 2) sum01 += dz;
  return add(x + dx[dir], y + dy[dir], dir ^ 1, dz);
}
void flip(int x, int y) {
  if (Free[x][y] == 0) {
    Free[x][y] = 1;
    Freecnt++;
    for (int dd = 0; dd < 4; dd++) {
      int cc = cnt(x, y, dd);
      add(x, y, dd ^ 3, cc);
    }
  } else {
    for (int dd = 0; dd < 4; dd++) {
      int cc = cnt(x, y, dd);
      add(x, y, dd ^ 3, -cc);
    }
    Free[x][y] = 0;
    Freecnt--;
  }
}
void init() {
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      Free[i][j] = 1;
      Freecnt++;
      dir[i][j][2] = dir[i][j][3] = 1;
      if (i) dir[i][j][2] += dir[i - 1][j][3];
      if (j) dir[i][j][3] += dir[i][j - 1][2];
    }
  }
  for (int i = n; i >= 1; i--) {
    for (int j = m; j >= 1; j--) {
      dir[i][j][0] = dir[i][j][1] = 1;
      if (i < n) dir[i][j][0] += dir[i + 1][j][1];
      if (j < m) dir[i][j][1] += dir[i][j + 1][0];
      sum01 += dir[i][j][0] + dir[i][j][1];
    }
  }
}
void all() {
  cout << sum01 << " " << Freecnt << " " << sum01 - Freecnt << endl;
  for (int d = 0; d < 4; d++) {
    cout << d << " -----> " << endl;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        cout << dir[i][j][d] << " ";
      }
      cout << endl;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  init();
  while (q--) {
    int x, y;
    cin >> x >> y;
    flip(x, y);
    cout << sum01 - Freecnt << "\n";
    cout.flush();
  }
}

```