#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long infinity = 1000000000000000000;
const int inf = 1e9 + 5;
const int maxn = 1005;
int n, m, q;
int state[maxn][maxn] = {0};
long long dp[maxn][maxn][2] = {0};
long long ans = 0;
int in(int x, int y) {
  if (x >= 0 && x < n && y >= 0 && y < m) {
    return 1;
  }
  return 0;
}
void g(int x, int y) {
  if (in(x + 1, y)) {
    ans -= dp[x][y][1];
    int extra = state[x + 1][y];
    dp[x][y][1] = extra + dp[x + 1][y][0];
    ans += dp[x][y][1];
  }
  if (in(x, y + 1)) {
    ans -= dp[x][y][0];
    int extra = state[x][y + 1];
    dp[x][y][0] = extra + dp[x][y + 1][1];
    ans += dp[x][y][0];
  }
}
void f(int x, int y) {
  if (!in(x, y)) {
    return;
  }
  if (!state[x][y]) {
    return;
  }
  g(x, y);
  if (in(x - 1, y) && state[x - 1][y]) {
    g(x - 1, y);
  }
  if (in(x, y - 1) && state[x][y - 1]) {
    g(x, y - 1);
  }
  f(x - 1, y - 1);
}
void solve() {
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      state[i][j] = 1;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      g(i, j);
      ans++;
    }
  }
  for (int Q = 0; Q < q; Q++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    state[x][y] ^= 1;
    if (!state[x][y]) {
      ans -= dp[x][y][0];
      ans -= dp[x][y][1];
      dp[x][y][0] = 0;
      dp[x][y][1] = 0;
      if (in(x - 1, y) && state[x - 1][y]) {
        g(x - 1, y);
      }
      if (in(x, y - 1) && state[x][y - 1]) {
        g(x, y - 1);
      }
      ans--;
      f(x - 1, y - 1);
    } else {
      ans++;
      f(x, y);
    }
    cout << ans << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int TC = 1;
  for (int tc = 1; tc <= TC; tc++) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
const long long mod = 1000000007;
const long long infinity = 1000000000000000000;
const int inf = 1e9 + 5;
const int maxn = 1005;
int n, m, q;
int state[maxn][maxn] = {0};
long long dp[maxn][maxn][2] = {0};
long long ans = 0;
int in(int x, int y) {
  if (x >= 0 && x < n && y >= 0 && y < m) {
    return 1;
  }
  return 0;
}
void g(int x, int y) {
  if (in(x + 1, y)) {
    ans -= dp[x][y][1];
    int extra = state[x + 1][y];
    dp[x][y][1] = extra + dp[x + 1][y][0];
    ans += dp[x][y][1];
  }
  if (in(x, y + 1)) {
    ans -= dp[x][y][0];
    int extra = state[x][y + 1];
    dp[x][y][0] = extra + dp[x][y + 1][1];
    ans += dp[x][y][0];
  }
}
void f(int x, int y) {
  if (!in(x, y)) {
    return;
  }
  if (!state[x][y]) {
    return;
  }
  g(x, y);
  if (in(x - 1, y) && state[x - 1][y]) {
    g(x - 1, y);
  }
  if (in(x, y - 1) && state[x][y - 1]) {
    g(x, y - 1);
  }
  f(x - 1, y - 1);
}
void solve() {
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      state[i][j] = 1;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = m - 1; j >= 0; j--) {
      g(i, j);
      ans++;
    }
  }
  for (int Q = 0; Q < q; Q++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    state[x][y] ^= 1;
    if (!state[x][y]) {
      ans -= dp[x][y][0];
      ans -= dp[x][y][1];
      dp[x][y][0] = 0;
      dp[x][y][1] = 0;
      if (in(x - 1, y) && state[x - 1][y]) {
        g(x - 1, y);
      }
      if (in(x, y - 1) && state[x][y - 1]) {
        g(x, y - 1);
      }
      ans--;
      f(x - 1, y - 1);
    } else {
      ans++;
      f(x, y);
    }
    cout << ans << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int TC = 1;
  for (int tc = 1; tc <= TC; tc++) {
    solve();
  }
  return 0;
}

```