#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int mp[1010][1010];
long long dpR[1010][1010], dpD[1010][1010];
long long Ans = 0;
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = n; i >= 1; i--) {
    for (int j = m; j >= 1; j--) {
      dpR[i][j] = 1 + dpD[i][j + 1];
      dpD[i][j] = 1 + dpR[i + 1][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      Ans += dpR[i][j] + dpD[i][j] - 1;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) mp[i][j] = 1;
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (mp[x][y]) {
      Ans -= dpR[x][y] + dpD[x][y] - 1;
      int xx = x, yy = y;
      int step = 0;
      while (xx > 0 && yy > 0) {
        if (step & 1)
          xx--;
        else
          yy--;
        if (!mp[xx][yy]) break;
        if (xx <= 0 || yy <= 0) break;
        if (step & 1) {
          dpD[xx][yy] -= dpD[x][y];
          Ans -= dpD[x][y];
        } else {
          dpR[xx][yy] -= dpD[x][y];
          Ans -= dpD[x][y];
        }
        step++;
      }
      xx = x, yy = y;
      step = 0;
      while (xx > 0 && yy > 0) {
        if (step & 1)
          yy--;
        else
          xx--;
        if (!mp[xx][yy]) break;
        if (xx <= 0 || yy <= 0) break;
        if (step & 1) {
          dpR[xx][yy] -= dpR[x][y];
          Ans -= dpR[x][y];
        } else {
          dpD[xx][yy] -= dpR[x][y];
          Ans -= dpR[x][y];
        }
        step++;
      }
      dpR[x][y] = dpD[x][y] = 0;
      mp[x][y] = 0;
    } else {
      dpR[x][y] = dpD[x][y + 1] + 1;
      dpD[x][y] = dpR[x + 1][y] + 1;
      Ans += dpR[x][y] + dpD[x][y] - 1;
      int xx = x, yy = y;
      int step = 0;
      while (xx > 0 && yy > 0) {
        if (step & 1)
          xx--;
        else
          yy--;
        if (!mp[xx][yy]) break;
        if (xx <= 0 || yy <= 0) break;
        if (step & 1) {
          dpD[xx][yy] += dpD[x][y];
          Ans += dpD[x][y];
        } else {
          dpR[xx][yy] += dpD[x][y];
          Ans += dpD[x][y];
        }
        step++;
      }
      xx = x, yy = y;
      step = 0;
      while (xx > 0 && yy > 0) {
        if (step & 1)
          yy--;
        else
          xx--;
        if (!mp[xx][yy]) break;
        if (xx <= 0 || yy <= 0) break;
        if (step & 1) {
          dpR[xx][yy] += dpR[x][y];
          Ans += dpR[x][y];
        } else {
          dpD[xx][yy] += dpR[x][y];
          Ans += dpR[x][y];
        }
        step++;
      }
      mp[x][y] = 1;
    }
    printf("%lld\n", Ans);
  }
  return 0;
}

### Prompt

Please formulate 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;
int n, m, q;
int mp[1010][1010];
long long dpR[1010][1010], dpD[1010][1010];
long long Ans = 0;
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = n; i >= 1; i--) {
    for (int j = m; j >= 1; j--) {
      dpR[i][j] = 1 + dpD[i][j + 1];
      dpD[i][j] = 1 + dpR[i + 1][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      Ans += dpR[i][j] + dpD[i][j] - 1;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) mp[i][j] = 1;
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (mp[x][y]) {
      Ans -= dpR[x][y] + dpD[x][y] - 1;
      int xx = x, yy = y;
      int step = 0;
      while (xx > 0 && yy > 0) {
        if (step & 1)
          xx--;
        else
          yy--;
        if (!mp[xx][yy]) break;
        if (xx <= 0 || yy <= 0) break;
        if (step & 1) {
          dpD[xx][yy] -= dpD[x][y];
          Ans -= dpD[x][y];
        } else {
          dpR[xx][yy] -= dpD[x][y];
          Ans -= dpD[x][y];
        }
        step++;
      }
      xx = x, yy = y;
      step = 0;
      while (xx > 0 && yy > 0) {
        if (step & 1)
          yy--;
        else
          xx--;
        if (!mp[xx][yy]) break;
        if (xx <= 0 || yy <= 0) break;
        if (step & 1) {
          dpR[xx][yy] -= dpR[x][y];
          Ans -= dpR[x][y];
        } else {
          dpD[xx][yy] -= dpR[x][y];
          Ans -= dpR[x][y];
        }
        step++;
      }
      dpR[x][y] = dpD[x][y] = 0;
      mp[x][y] = 0;
    } else {
      dpR[x][y] = dpD[x][y + 1] + 1;
      dpD[x][y] = dpR[x + 1][y] + 1;
      Ans += dpR[x][y] + dpD[x][y] - 1;
      int xx = x, yy = y;
      int step = 0;
      while (xx > 0 && yy > 0) {
        if (step & 1)
          xx--;
        else
          yy--;
        if (!mp[xx][yy]) break;
        if (xx <= 0 || yy <= 0) break;
        if (step & 1) {
          dpD[xx][yy] += dpD[x][y];
          Ans += dpD[x][y];
        } else {
          dpR[xx][yy] += dpD[x][y];
          Ans += dpD[x][y];
        }
        step++;
      }
      xx = x, yy = y;
      step = 0;
      while (xx > 0 && yy > 0) {
        if (step & 1)
          yy--;
        else
          xx--;
        if (!mp[xx][yy]) break;
        if (xx <= 0 || yy <= 0) break;
        if (step & 1) {
          dpR[xx][yy] += dpR[x][y];
          Ans += dpR[x][y];
        } else {
          dpD[xx][yy] += dpR[x][y];
          Ans += dpR[x][y];
        }
        step++;
      }
      mp[x][y] = 1;
    }
    printf("%lld\n", Ans);
  }
  return 0;
}

```