#include <bits/stdc++.h>
using namespace std;
long long n, m, k;
long long read() {
  long long s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}
long long a[1010][1010];
long long dp[1010][1010][2];
signed main() {
  long long T = 1;
  while (T--) {
    long long n = read(), m = read(), q = read();
    for (long long i = 1, iend = n; i <= iend; ++i) {
      for (long long j = 1, jend = m; j <= jend; ++j) {
        a[i][j] = 1;
      }
    }
    long long ans = 0;
    for (long long i = 1, iend = n; i <= iend; ++i) {
      for (long long j = 1, jend = m; j <= jend; ++j) {
        dp[i][j][0] = dp[i][j][1] = 1;
        dp[i][j][0] += dp[i - 1][j][1];
        dp[i][j][1] += dp[i][j - 1][0];
        ans += dp[i][j][0] + dp[i][j][1] - 1;
      }
    }
    while (q--) {
      long long x = read(), y = read();
      long long nx = x, ny = y;
      long long lf = 0;
      while (1) {
        if (lf % 2 == 0) {
          --nx;
          if (a[nx][ny] == 0) break;
          lf++;
        } else {
          --ny;
          if (a[nx][ny] == 0) break;
          lf++;
        }
      }
      long long rf = 0;
      nx = x, ny = y;
      while (1) {
        if (rf % 2 == 1) {
          ++nx;
          if (a[nx][ny] == 0) break;
          rf++;
        } else {
          ++ny;
          if (a[nx][ny] == 0) break;
          rf++;
        }
      }
      long long sum1 = (rf + 1) * (lf + 1);
      nx = x, ny = y;
      lf = 0;
      while (1) {
        if (lf % 2 == 1) {
          --nx;
          if (a[nx][ny] == 0) break;
          lf++;
        } else {
          --ny;
          if (a[nx][ny] == 0) break;
          lf++;
        }
      }
      rf = 0;
      nx = x, ny = y;
      while (1) {
        if (rf % 2 == 0) {
          ++nx;
          if (a[nx][ny] == 0) break;
          rf++;
        } else {
          ++ny;
          if (a[nx][ny] == 0) break;
          rf++;
        }
      }
      long long sum2 = (rf + 1) * (lf + 1);
      a[x][y] = 1 - a[x][y];
      sum1 += sum2;
      sum1--;
      if (a[x][y] == 1)
        ans += sum1;
      else
        ans -= sum1;
      cout << ans << '\n';
    }
  }
}

### Prompt

Develop a solution in cpp 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;
long long n, m, k;
long long read() {
  long long s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}
long long a[1010][1010];
long long dp[1010][1010][2];
signed main() {
  long long T = 1;
  while (T--) {
    long long n = read(), m = read(), q = read();
    for (long long i = 1, iend = n; i <= iend; ++i) {
      for (long long j = 1, jend = m; j <= jend; ++j) {
        a[i][j] = 1;
      }
    }
    long long ans = 0;
    for (long long i = 1, iend = n; i <= iend; ++i) {
      for (long long j = 1, jend = m; j <= jend; ++j) {
        dp[i][j][0] = dp[i][j][1] = 1;
        dp[i][j][0] += dp[i - 1][j][1];
        dp[i][j][1] += dp[i][j - 1][0];
        ans += dp[i][j][0] + dp[i][j][1] - 1;
      }
    }
    while (q--) {
      long long x = read(), y = read();
      long long nx = x, ny = y;
      long long lf = 0;
      while (1) {
        if (lf % 2 == 0) {
          --nx;
          if (a[nx][ny] == 0) break;
          lf++;
        } else {
          --ny;
          if (a[nx][ny] == 0) break;
          lf++;
        }
      }
      long long rf = 0;
      nx = x, ny = y;
      while (1) {
        if (rf % 2 == 1) {
          ++nx;
          if (a[nx][ny] == 0) break;
          rf++;
        } else {
          ++ny;
          if (a[nx][ny] == 0) break;
          rf++;
        }
      }
      long long sum1 = (rf + 1) * (lf + 1);
      nx = x, ny = y;
      lf = 0;
      while (1) {
        if (lf % 2 == 1) {
          --nx;
          if (a[nx][ny] == 0) break;
          lf++;
        } else {
          --ny;
          if (a[nx][ny] == 0) break;
          lf++;
        }
      }
      rf = 0;
      nx = x, ny = y;
      while (1) {
        if (rf % 2 == 0) {
          ++nx;
          if (a[nx][ny] == 0) break;
          rf++;
        } else {
          ++ny;
          if (a[nx][ny] == 0) break;
          rf++;
        }
      }
      long long sum2 = (rf + 1) * (lf + 1);
      a[x][y] = 1 - a[x][y];
      sum1 += sum2;
      sum1--;
      if (a[x][y] == 1)
        ans += sum1;
      else
        ans -= sum1;
      cout << ans << '\n';
    }
  }
}

```