#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
long long n, m, i, j, k, q;
long long ans, x, y;
long long a[1010][1010];
long long u1, u2, v1, v2;
signed main() {
  n = read();
  m = read();
  q = read();
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= m; ++j) ans += (min(i - 1, j - 1) * 4 + 1 + (i != j));
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= m; ++j) a[i][j] = 1;
  for (i = 0; i <= n + 1; ++i) a[i][0] = a[i][m + 1] = -1;
  for (i = 0; i <= m + 1; ++i) a[0][i] = a[n + 1][i] = -1;
  while (q--) {
    x = read();
    y = read();
    a[x][y] = -a[x][y];
    ans += a[x][y];
    u1 = v1 = 0;
    u2 = v2 = 0;
    for (i = x, j = y;;) {
      --i;
      if (a[i][j] == -1) break;
      ++u1;
      --j;
      if (a[i][j] == -1) break;
      ++u1;
    }
    for (i = x, j = y;;) {
      --j;
      if (a[i][j] == -1) break;
      ++u2;
      --i;
      if (a[i][j] == -1) break;
      ++u2;
    }
    for (i = x, j = y;;) {
      ++i;
      if (a[i][j] == -1) break;
      ++v1;
      ++j;
      if (a[i][j] == -1) break;
      ++v1;
    }
    for (i = x, j = y;;) {
      ++j;
      if (a[i][j] == -1) break;
      ++v2;
      ++i;
      if (a[i][j] == -1) break;
      ++v2;
    }
    ans += a[x][y] * (u1 + v1 + u2 + v2 + u1 * v2 + u2 * v1);
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
long long n, m, i, j, k, q;
long long ans, x, y;
long long a[1010][1010];
long long u1, u2, v1, v2;
signed main() {
  n = read();
  m = read();
  q = read();
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= m; ++j) ans += (min(i - 1, j - 1) * 4 + 1 + (i != j));
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= m; ++j) a[i][j] = 1;
  for (i = 0; i <= n + 1; ++i) a[i][0] = a[i][m + 1] = -1;
  for (i = 0; i <= m + 1; ++i) a[0][i] = a[n + 1][i] = -1;
  while (q--) {
    x = read();
    y = read();
    a[x][y] = -a[x][y];
    ans += a[x][y];
    u1 = v1 = 0;
    u2 = v2 = 0;
    for (i = x, j = y;;) {
      --i;
      if (a[i][j] == -1) break;
      ++u1;
      --j;
      if (a[i][j] == -1) break;
      ++u1;
    }
    for (i = x, j = y;;) {
      --j;
      if (a[i][j] == -1) break;
      ++u2;
      --i;
      if (a[i][j] == -1) break;
      ++u2;
    }
    for (i = x, j = y;;) {
      ++i;
      if (a[i][j] == -1) break;
      ++v1;
      ++j;
      if (a[i][j] == -1) break;
      ++v1;
    }
    for (i = x, j = y;;) {
      ++j;
      if (a[i][j] == -1) break;
      ++v2;
      ++i;
      if (a[i][j] == -1) break;
      ++v2;
    }
    ans += a[x][y] * (u1 + v1 + u2 + v2 + u1 * v2 + u2 * v1);
    printf("%lld\n", ans);
  }
  return 0;
}

```