#include <bits/stdc++.h>
#pragma GCC optimize(2, 3, "Ofast")
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
long long can[1005][1005];
int a[1005][1005];
int n, m, q;
long long ans;
int main() {
  n = read(), m = read(), q = read();
  for (int i = n; i >= 1; i--) {
    for (int j = m; j >= 1; j--) {
      can[i][j] = can[i + 1][j + 1] + 1 + (i < n) + (j < m) + (i < n && j < m);
      ans += can[i][j];
    }
  }
  while (q--) {
    int x = read(), y = read();
    if (a[x][y]) {
      a[x][y] = 0;
      int x0 = 0, y0 = 0;
      {
        int zx = x, zy = y;
        int dir = 0;
        while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
          x0++;
          if (dir)
            zx--;
          else
            zy--;
          dir ^= 1;
        }
      }
      {
        int zx = x, zy = y;
        int dir = 1;
        while (zx <= n && zy <= m && !a[zx][zy]) {
          y0++;
          if (dir)
            zx++;
          else
            zy++;
          dir ^= 1;
        }
      }
      ans += 1ll * x0 * y0;
      x0 = 0, y0 = 0;
      {
        int zx = x, zy = y;
        int dir = 1;
        while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
          x0++;
          if (dir)
            zx--;
          else
            zy--;
          dir ^= 1;
        }
      }
      {
        int zx = x, zy = y;
        int dir = 0;
        while (zx <= n && zy <= m && !a[zx][zy]) {
          y0++;
          if (dir)
            zx++;
          else
            zy++;
          dir ^= 1;
        }
      }
      ans += 1ll * x0 * y0 - 1;
    } else {
      int x0 = 0, y0 = 0;
      {
        int zx = x, zy = y;
        int dir = 0;
        while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
          x0++;
          if (dir)
            zx--;
          else
            zy--;
          dir ^= 1;
        }
      }
      {
        int zx = x, zy = y;
        int dir = 1;
        while (zx <= n && zy <= m && !a[zx][zy]) {
          y0++;
          if (dir)
            zx++;
          else
            zy++;
          dir ^= 1;
        }
      }
      ans -= 1ll * x0 * y0;
      x0 = 0, y0 = 0;
      {
        int zx = x, zy = y;
        int dir = 1;
        while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
          x0++;
          if (dir)
            zx--;
          else
            zy--;
          dir ^= 1;
        }
      }
      {
        int zx = x, zy = y;
        int dir = 0;
        while (zx <= n && zy <= m && !a[zx][zy]) {
          y0++;
          if (dir)
            zx++;
          else
            zy++;
          dir ^= 1;
        }
      }
      ans -= 1ll * x0 * y0 - 1;
      a[x][y] = 1;
    }
    print(ans, '\n');
  }
}

### 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>
#pragma GCC optimize(2, 3, "Ofast")
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
long long can[1005][1005];
int a[1005][1005];
int n, m, q;
long long ans;
int main() {
  n = read(), m = read(), q = read();
  for (int i = n; i >= 1; i--) {
    for (int j = m; j >= 1; j--) {
      can[i][j] = can[i + 1][j + 1] + 1 + (i < n) + (j < m) + (i < n && j < m);
      ans += can[i][j];
    }
  }
  while (q--) {
    int x = read(), y = read();
    if (a[x][y]) {
      a[x][y] = 0;
      int x0 = 0, y0 = 0;
      {
        int zx = x, zy = y;
        int dir = 0;
        while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
          x0++;
          if (dir)
            zx--;
          else
            zy--;
          dir ^= 1;
        }
      }
      {
        int zx = x, zy = y;
        int dir = 1;
        while (zx <= n && zy <= m && !a[zx][zy]) {
          y0++;
          if (dir)
            zx++;
          else
            zy++;
          dir ^= 1;
        }
      }
      ans += 1ll * x0 * y0;
      x0 = 0, y0 = 0;
      {
        int zx = x, zy = y;
        int dir = 1;
        while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
          x0++;
          if (dir)
            zx--;
          else
            zy--;
          dir ^= 1;
        }
      }
      {
        int zx = x, zy = y;
        int dir = 0;
        while (zx <= n && zy <= m && !a[zx][zy]) {
          y0++;
          if (dir)
            zx++;
          else
            zy++;
          dir ^= 1;
        }
      }
      ans += 1ll * x0 * y0 - 1;
    } else {
      int x0 = 0, y0 = 0;
      {
        int zx = x, zy = y;
        int dir = 0;
        while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
          x0++;
          if (dir)
            zx--;
          else
            zy--;
          dir ^= 1;
        }
      }
      {
        int zx = x, zy = y;
        int dir = 1;
        while (zx <= n && zy <= m && !a[zx][zy]) {
          y0++;
          if (dir)
            zx++;
          else
            zy++;
          dir ^= 1;
        }
      }
      ans -= 1ll * x0 * y0;
      x0 = 0, y0 = 0;
      {
        int zx = x, zy = y;
        int dir = 1;
        while (zx >= 1 && zy >= 1 && !a[zx][zy]) {
          x0++;
          if (dir)
            zx--;
          else
            zy--;
          dir ^= 1;
        }
      }
      {
        int zx = x, zy = y;
        int dir = 0;
        while (zx <= n && zy <= m && !a[zx][zy]) {
          y0++;
          if (dir)
            zx++;
          else
            zy++;
          dir ^= 1;
        }
      }
      ans -= 1ll * x0 * y0 - 1;
      a[x][y] = 1;
    }
    print(ans, '\n');
  }
}

```