#include <bits/stdc++.h>
using namespace std;
int q, n, m;
struct node {
  int x, y;
};
queue<node> Q;
int a[1005][1005];
int vis[1005][1005];
int use, nuse;
int dp[1005][1005][2];
int bfs(int x, int y, int op, int z) {
  int qwq = 0;
  Q.push({x, y});
  while (Q.size()) {
    node v = Q.front();
    Q.pop();
    if (op) {
      v.x += z;
      op ^= 1;
    } else {
      v.y += z;
      op ^= 1;
    }
    if (v.x >= 1 && v.x <= n && v.y >= 1 && v.y <= m) {
      if (!vis[v.x][v.y]) {
        qwq++;
        Q.push({v.x, v.y});
      } else
        return qwq;
    }
  }
  return qwq;
}
int main() {
  long long ans = 0;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp[i][j][0] = dp[i][j][1] = 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp[i][j][0] += dp[i][j - 1][1];
      dp[i][j][1] += dp[i - 1][j][0];
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp[i][j][0]--;
      dp[i][j][1]--;
      ans += dp[i][j][0] + dp[i][j][1] + 1;
    }
  }
  int x, y;
  long long now = ans;
  while (q--) {
    scanf("%d%d", &x, &y);
    int a1 = bfs(x, y, 0, 1);
    int a2 = bfs(x, y, 1, 1);
    int b1 = bfs(x, y, 0, -1);
    int b2 = bfs(x, y, 1, -1);
    if (!vis[x][y]) {
      now -= ((a1 * b2) + (a2 * b1) + a1 + b1 + a2 + b2);
      now--;
      vis[x][y] = 1;
      printf("%lld\n", now);
    } else {
      now += (a1 * b2) + (a2 * b1) + a1 + b1 + a2 + b2;
      now++;
      vis[x][y] = 0;
      printf("%lld\n", now);
    }
  }
  return 0;
}

### 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;
int q, n, m;
struct node {
  int x, y;
};
queue<node> Q;
int a[1005][1005];
int vis[1005][1005];
int use, nuse;
int dp[1005][1005][2];
int bfs(int x, int y, int op, int z) {
  int qwq = 0;
  Q.push({x, y});
  while (Q.size()) {
    node v = Q.front();
    Q.pop();
    if (op) {
      v.x += z;
      op ^= 1;
    } else {
      v.y += z;
      op ^= 1;
    }
    if (v.x >= 1 && v.x <= n && v.y >= 1 && v.y <= m) {
      if (!vis[v.x][v.y]) {
        qwq++;
        Q.push({v.x, v.y});
      } else
        return qwq;
    }
  }
  return qwq;
}
int main() {
  long long ans = 0;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp[i][j][0] = dp[i][j][1] = 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp[i][j][0] += dp[i][j - 1][1];
      dp[i][j][1] += dp[i - 1][j][0];
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp[i][j][0]--;
      dp[i][j][1]--;
      ans += dp[i][j][0] + dp[i][j][1] + 1;
    }
  }
  int x, y;
  long long now = ans;
  while (q--) {
    scanf("%d%d", &x, &y);
    int a1 = bfs(x, y, 0, 1);
    int a2 = bfs(x, y, 1, 1);
    int b1 = bfs(x, y, 0, -1);
    int b2 = bfs(x, y, 1, -1);
    if (!vis[x][y]) {
      now -= ((a1 * b2) + (a2 * b1) + a1 + b1 + a2 + b2);
      now--;
      vis[x][y] = 1;
      printf("%lld\n", now);
    } else {
      now += (a1 * b2) + (a2 * b1) + a1 + b1 + a2 + b2;
      now++;
      vis[x][y] = 0;
      printf("%lld\n", now);
    }
  }
  return 0;
}

```