#include <bits/stdc++.h>
using namespace std;
inline void debugMode() {}
long long dp1[1001][1001], dp2[1001][1001];
int state[1001][1001];
void solvecases(int cse) {
  long long n, m, q;
  cin >> n >> m >> q;
  long long cnt = 1ll * n * m;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp1[i][j] = dp2[i - 1][j] + 1;
      dp2[i][j] = dp1[i][j - 1] + 1;
      ans += dp1[i][j] + dp2[i][j];
    }
  }
  ++n;
  ++m;
  for (int i = 0; i < q; ++i) {
    int x, y;
    cin >> x >> y;
    long long a, b;
    if (state[x][y]) {
      a = dp2[x - 1][y] + 1;
      b = dp1[x][y - 1] + 1;
      state[x][y] = 0;
      ++cnt;
      dp1[x][y] = a;
      dp2[x][y] = b;
      ans += a + b;
      if (((x) < n && (y + 1) < m && (x) >= 0 && (y + 1) >= 0) &&
          !state[x][y + 1]) {
        dp2[x][y + 1] += a;
        ans += a;
      } else
        a = 0;
    } else {
      a = -dp1[x][y];
      b = -dp2[x][y];
      state[x][y] = 1;
      --cnt;
      dp1[x][y] = dp2[x][y] = 0;
      ans += a + b;
      if (((x) < n && (y + 1) < m && (x) >= 0 && (y + 1) >= 0) &&
          !state[x][y + 1]) {
        dp2[x][y + 1] += a;
        ans += a;
      } else
        a = 0;
    }
    int j, k;
    j = x + 1;
    k = y + 1;
    while (j < n && k < m) {
      if (state[j][k - 1]) {
        b = 0;
      }
      if (state[j][k]) a = 0;
      dp1[j][k - 1] += b;
      dp1[j][k] += a;
      ans += a + b;
      if (state[j][k]) {
        b = 0;
      }
      if (((j) < n && (k + 1) < m && (j) >= 0 && (k + 1) >= 0)) {
        if (state[j][k + 1]) a = 0;
        dp2[j][k + 1] += a;
        ans += a;
      }
      dp2[j][k] += b;
      ans += b;
      ++j;
      ++k;
    }
    if (j < n) {
      if (!state[j][k - 1]) {
        dp1[j][k - 1] += b;
        ans += b;
      }
    }
    cout << ans - cnt << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solvecases(i);
  }
}

### Prompt

Please create a solution in cpp to the following problem:

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;
inline void debugMode() {}
long long dp1[1001][1001], dp2[1001][1001];
int state[1001][1001];
void solvecases(int cse) {
  long long n, m, q;
  cin >> n >> m >> q;
  long long cnt = 1ll * n * m;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp1[i][j] = dp2[i - 1][j] + 1;
      dp2[i][j] = dp1[i][j - 1] + 1;
      ans += dp1[i][j] + dp2[i][j];
    }
  }
  ++n;
  ++m;
  for (int i = 0; i < q; ++i) {
    int x, y;
    cin >> x >> y;
    long long a, b;
    if (state[x][y]) {
      a = dp2[x - 1][y] + 1;
      b = dp1[x][y - 1] + 1;
      state[x][y] = 0;
      ++cnt;
      dp1[x][y] = a;
      dp2[x][y] = b;
      ans += a + b;
      if (((x) < n && (y + 1) < m && (x) >= 0 && (y + 1) >= 0) &&
          !state[x][y + 1]) {
        dp2[x][y + 1] += a;
        ans += a;
      } else
        a = 0;
    } else {
      a = -dp1[x][y];
      b = -dp2[x][y];
      state[x][y] = 1;
      --cnt;
      dp1[x][y] = dp2[x][y] = 0;
      ans += a + b;
      if (((x) < n && (y + 1) < m && (x) >= 0 && (y + 1) >= 0) &&
          !state[x][y + 1]) {
        dp2[x][y + 1] += a;
        ans += a;
      } else
        a = 0;
    }
    int j, k;
    j = x + 1;
    k = y + 1;
    while (j < n && k < m) {
      if (state[j][k - 1]) {
        b = 0;
      }
      if (state[j][k]) a = 0;
      dp1[j][k - 1] += b;
      dp1[j][k] += a;
      ans += a + b;
      if (state[j][k]) {
        b = 0;
      }
      if (((j) < n && (k + 1) < m && (j) >= 0 && (k + 1) >= 0)) {
        if (state[j][k + 1]) a = 0;
        dp2[j][k + 1] += a;
        ans += a;
      }
      dp2[j][k] += b;
      ans += b;
      ++j;
      ++k;
    }
    if (j < n) {
      if (!state[j][k - 1]) {
        dp1[j][k - 1] += b;
        ans += b;
      }
    }
    cout << ans - cnt << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solvecases(i);
  }
}

```